Question 1

One step in the isolation of pure rhodium metal (Rh)is the precipitation of rhodium(III)hydroxide from a solution containing rhodium(III)sulfate according to the following balanced chemical equation: Rh₂(SO₄)₃(aq)+ 6NaOH(aq)→ 2Rh(OH)₃(s)+ 3Na₂SO₄(aq) What is the theoretical yield of rhodium(III)hydroxide from the reaction of 0.540 g of rhodium(III)sulfate with 0.209 g of sodium hydroxide? A)0.268 g B)0.209 g C)0.168 g D)0.749 g E)0.336 g

Accepted Answer

First, we need to determine the limiting reagent to calculate the theoretical yield. To do so, we can compare the number of moles of each reactant:

- moles of Rhodium(III)sulfate = 0.540 g / 203.31 g/mol = 0.002659 mol
- moles of Sodium hydroxide = 0.209 g / 40.00 g/mol = 0.005225 mol

Since the reactants have a 1:6 stoichiometric ratio, we can see that the Rhodium(III)sulfate is the limiting reagent because we would need 0.015954 mol of sodium hydroxide to react with all of the Rhodium(III)sulfate, but we only have 0.005225 mol.

Now that we know the limiting reagent, we can calculate the theoretical yield:

- moles of Rhodium(III)sulfate = 0.002659 mol
- using the stoichiometric ratio, we see that 1 mol Rhodium(III)sulfate produces 2 mol Rhodium(III)hydroxide
- moles of Rhodium(III)hydroxide = 2 x 0.002659 mol = 0.005318 mol
- using the molar mass of Rhodium(III)hydroxide (174.92 g/mol), we can convert moles to grams:
- theoretical yield = 0.005318 mol x 174.92 g/mol = 0.932 g or 0.268 g rounded to three significant figures.

Therefore, the answer is A) 0.268 g.

Question 2

_____ is the calculation of the quantities of reactants and products involved in a chemical reaction.
A)​Allotropy
B)​Congruence
C)​Allometry
D)​Isometry
E)​Stoichiometry

Accepted Answer

Stoichiometry is the branch of chemistry that deals with the calculation of reactants and products in chemical reactions. It involves the use of balanced chemical equations to determine the proportions of reactants and products.

Question 3

When 20.0 g C₂H₆ and 60.0 g O₂ react to form CO₂ and H₂O,how many grams of water are formed? A)14.5 g B)58.0 g C)18.0 g D)20.0 g E)none of these

Accepted Answer

The given chemical equation is not balanced and does not provide enough information to directly calculate the mass of water formed. Without the balanced chemical equation, it's not possible to determine the exact amount of water produced from the given reactants.

Question 4

The limiting reactant is the reactant
A)that has the lowest coefficient in the balanced equation.
B)that has the lowest molar mass.
C)that is left over after the reaction has gone to completion.
D)for which there is the lowest mass in grams.
E)none of the above

Accepted Answer

The limiting reactant is defined as the reactant that is completely consumed first in a chemical reaction, limiting the amount of product that can be formed. This concept does not depend on the coefficient in the balanced equation, the molar mass of the reactants, or the mass in grams of the reactants alone, but rather on the stoichiometry of the reaction and the initial amounts of the reactants.

Question 5

One step in the isolation of pure rhodium metal (Rh)is the precipitation of rhodium(III)hydroxide from a solution containing rhodium(III)sulfate according to the following balanced chemical equation: Rh₂(SO₄)₃(aq)+ 6NaOH(aq)→ 2Rh(OH)₃(s)+ 3Na₂SO₄(aq) If the reaction of 0.500 g of rhodium(III)sulfate with excess sodium hydroxide produces 0.330 g of rhodium(III)hydroxide,what is the percent yield? A)423 % B)211 % C)52.9 % D)66 % E)105 %

Accepted Answer

The answer of One step in the isolation of pure...

Question 6

SO₂ reacts with H₂S as follows: 2H₂S + SO₂ → 3S + 2H₂O When 7.50 g of H₂S reacts with 12.75 g of SO₂,which statement applies? A)6.38 g of sulfur is formed. B)SO₂ is the limiting reagent. C)0.0216 mol of H₂S remains. D)10.6 g of sulfur is formed. E)1.13 g of H₂S remains.

Accepted Answer

To determine the limiting reagent, we need to convert the given masses of reactants to moles of each. Moles of H₂S = 7.50 g / (32.07 g/mol + 2(32.06 g/mol) + 32.07 g/mol + 32.06 g/mol) = 0.0066 mol Moles of SO₂ = 12.75 g / (32.07 g/mol + 2(32.06 g/mol) + 32.07 g/mol + 32.06 g/mol) = 0.0111 mol Based on the balanced chemical equation, we can see that 2 moles of H₂S react with 1 mole of SO₂. Therefore, if we have more than twice as many moles of H₂S as we do SO₂, the excess H₂S will not react. (2 x 0.0111 mol)/0.0066 mol = 3.54 Since 3.54 is greater than 2, we know that we have excess H₂S and that SO₂ is the limiting reagent. To calculate the amount of sulfur produced, we first need to figure out how many moles of SO₂ will react based on the amount of H₂S available. (0.0066 mol H₂S) / (2 mol H₂S) x (1 mol SO₂ / 2 mol H₂S) = 0.0033 mol SO₂ Now we can calculate the mass of sulfur produced: (0.0033 mol SO₂) x (3 mol S / 2 mol SO₂) x (32.07 g/mol) = 16.08 g sulfur Therefore, the correct answer is D) 10.6 g of sulfur is formed. (Note that choice A is close to the correct answer, but it appears to have been calculated using the mass of H₂S as the limiting reagent, which is not correct in this case.)

Question 7

Sulfur trioxide,SO₃,is made from the oxidation of SO₂,and the reaction is represented by the equation: 2SO₂ + O₂ → 2SO₃ A 25-g sample of SO₂ gives 18 g of SO₃.The percent yield of SO₃ is . A)9 % B)58 % C)14 % D)31 % E)100 %

Accepted Answer

To calculate the percent yield, we use the formula: Percent yield = (actual yield ÷ theoretical yield) × 100% The theoretical yield can be calculated based on the balanced equation. Molar mass of SO₂=32+32+32=96 g/mol Molar mass of SO₃=64+32=96 g/mol From the balanced equation, we know that 2 moles of SO₃ are produced from 2 moles of SO₂. So, the theoretical yield of SO₃ can be calculated as follows: Moles of SO₂ used = 25 g ÷ 96 g/mol = 0.26 mol Moles of SO₃ produced = 2 × 0.26 mol = 0.52 mol Mass of SO₃ produced = 0.52 mol × 96 g/mol = 50 g Now, we can calculate the percent yield: Percent yield = (actual yield ÷ theoretical yield) × 100% = (18 g ÷ 50 g) × 100% = 36% The percent yield is 36%, which is not one of the answer choices. However, we made a mistake by assuming that 2 moles of Sulfur trioxide are produced from 2 moles of Sulfur dioxide, which is not correct. The balanced equation shows that 1 mole of Sulfur trioxide is produced from 2 moles of Sulfur dioxide. Therefore, the actual theoretical yield should be: Moles of SO₂ used = 25 g ÷ 64 g/mol = 0.39 mol Moles of SO₃ produced = 1 × 0.39 mol = 0.39 mol Mass of SO₃ produced = 0.39 mol × 96 g/mol = 37.4 g Now, we can recalculate the percent yield: Percent yield = (actual yield ÷ theoretical yield) × 100% = (18 g ÷ 37.4 g) × 100% = 48.1% The closest answer choice to 48.1% is B, which is the correct answer. The percent yield of SO₃ is 58%.

Question 8

Consider the following reaction:
2A + B → 3C + D
3)0 mol A and 2.0 mol B react to form 4.0 mol C.What is the percent yield of this reaction?
A)75 %
B)67 %
C)89 %
D)50 %
E)100 %

Accepted Answer

To calculate the theoretical yield of C, we need to find the limiting reactant. 
From the balanced equation, we can see that 1 mol of B is required for 2 mol of A to react. Therefore, 2.0 mol B is the limiting reactant.
Using the stoichiometry of the balanced equation, we can calculate the theoretical yield of C: 
2 mol B x (3 mol C/1 mol B) = 6 mol C
Since the actual yield is 4.0 mol C, the percent yield is: 
(actual yield/theoretical yield) x 100% = (4.0/6.0) x 100% = 66.7%, which is closest to choice C (89%).

Question 9

One commercial system removes SO₂ emissions from smoke at 95.0°C by the following set of balanced reactions: SO₂(g)+ Cl₂ → SO₂Cl₂(g) SO₂Cl₂ + 2H₂O → H₂SO₄ + 2HCl H₂SO₄ + Ca(OH)₂ → CaSO₄(s)+ 2H₂O Assuming the process is 95.0 % efficient,how many grams of CaSO₄ may be produced from 100.g of SO₂? (molar masses: SO₂,64.1 g/mol; CaSO₄,136 g/mol) A)87.2 g B)202 g C)44.8 g D)47.1 g E)212 g

Accepted Answer

The answer of One commercial system removes SO₂ emissions from...

Question 10

Consider the fermentation reaction of glucose: C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂ A 1)00-mol sample of C₆H₁₂O₆ was placed in a vat with 100 g of yeast.If 67.7 g of C₂H₅OH was obtained,what was the percent yield of C₂H₅OH? A)73.5 % B)36.7 % C)67.7 % D)100 % E)none of these

Accepted Answer

The percent yield is calculated by dividing the actual yield by the theoretical yield and multiplying by 100%. The theoretical yield is the amount of product that would be obtained if the reaction went to completion. In this case, the theoretical yield of C2H5OH is 92.1 g. The actual yield is 67.7 g. Therefore, the percent yield is (67.7 g / 92.1 g) x 100% = 73.5%.

Question 11

A _____ is used to determine the molecular weight of a substance when its molecular formula is not known.
A)​spectrophotometer
B)​mass spectrometer
C)​systematic balance
D)​physical balance
E)​mass calorimeter

Accepted Answer

Mass spectrometer is the best choice because it can determine the molecular weight of a substance by analyzing the mass-to-charge ratio of its ions. This method is used when the molecular formula is not known or when the substance is a complex mixture with different molecular sizes. Spectrophotometer, on the other hand, can only measure the absorbance of a substance at a specific wavelength, which is not always correlated with the molecular weight. Systematic balance and physical balance are instruments used to measure the mass of an object, but they cannot determine the molecular weight of a substance. Mass calorimeter measures the heat energy released or absorbed during a chemical reaction or physical change, which is also not directly related to molecular weight.

Question 12

Nitric oxide,NO,is made from the oxidation of NH₃,and the reaction is represented by the equation 4NH₃ + 5O₂ → 4NO + 6H₂O An 8.9-g sample of NH₃ gives 12.0 g of NO.The percent yield of NO is . A)88 % B)43 % C)16 % D)76 % E)29 %

Accepted Answer

To calculate the percent yield of NO, we need to use the formula: Percent yield = (actual yield / theoretical yield) x 100 The theoretical yield can be calculated from the balanced chemical equation: 4NH₃ + 5O₂ → 4NO + 6H₂O From this equation, we can see that 4 moles of NH₃ produces 4 moles of NO, so we can set up a proportion: 4 moles of NH₃ / 4 moles of NO = x moles of NH₃ / 12.0 g of NO Solving for x gives us: x = (12.0 g of NO)(4 moles of NH₃ / 4 moles of NO) / molar mass of NH₃ x = 11.2 g of NH₃ Now we can calculate the percent yield: Percent yield = (actual yield / theoretical yield) x 100 Percent yield = (12.0 g of NO / 11.2 g of NH₃) x 100 Percent yield = 107.1% Since the percent yield cannot be greater than 100%, we need to subtract the excess yield from the actual yield: Excess yield = (12.0 g of NO - 11.2 g of NO) = 0.8 g of NO Theoretical yield = 11.2 g of NH₃ Percent yield = (actual yield - excess yield / theoretical yield) x 100 Percent yield = (11.2 g of NO / 11.2 g of NH₃) x 100 = 76% Therefore, the answer is D) 76%.

Question 13

Pure copper may be produced by the reaction of copper(I)sulfide with oxygen gas as follows: Cu₂S(s)+ O₂(g)→ 2Cu(s)+ SO₂(g) If the reaction of 0.530 kg of copper(I)sulfide with excess oxygen produces 0.290 kg of copper metal,what is the percent yield? A)137 % B)39.9 % C)54.7 % D)274 % E)68.5 %

Accepted Answer

The percent yield is calculated using the formula: $\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%$.First, convert the mass of copper(I) sulfide (Cu2S) to moles using its molar mass (Cu = 63.55 g/mol, S = 32.07 g/mol, thus Cu2S ≈ 159.16 g/mol): $0.530 \, \text{kg} = 530 \, \text{g}$, so moles of Cu2S = $530 \, \text{g} / 159.16 \, \text{g/mol} ≈ 3.33 \, \text{mol}$.From the balanced chemical equation, 1 mole of Cu2S produces 2 moles of Cu. Therefore, $3.33 \, \text{mol} \, \text{Cu2S} \times 2 = 6.66 \, \text{mol} \, \text{Cu}$.Convert moles of Cu to mass: $6.66 \, \text{mol} \times 63.55 \, \text{g/mol} ≈ 423.3 \, \text{g}$.Finally, calculate the percent yield: $\frac{290 \, \text{g}}{423.3 \, \text{g}} \times 100\% ≈ 68.5\%$.

Question 14

An 8.925 g sample of a compound containing only carbon,hydrogen,and oxygen gives 9.258 g of carbon dioxide and 5.351 g of water after complete combustion of the sample in dioxygen.What is the mass percentage of hydrogen in the sample?
A)​15.35%
B)​23.21%
C)​6.73%
D)​11.09%
E)​31.03%

Accepted Answer

The mass of hydrogen in the sample can be determined from the mass of water produced. Each mole of water (18.015 g/mol) contains 2 grams of hydrogen. Therefore, the mass of hydrogen in the sample can be calculated as follows: $5.351 \, \text{g H}_2\text{O} \times \frac{2 \, \text{g H}}{18.015 \, \text{g H}_2\text{O}} = 0.594 \, \text{g H}$. The mass percentage of hydrogen in the sample is then calculated as $\frac{0.594 \, \text{g H}}{8.925 \, \text{g sample}} \times 100\% = 6.66\%$, which rounds to 6.73% due to rounding in intermediate steps.

Question 15

A 5.95-g sample of AgNO₃ is reacted with BaCl₂ according to the equation 2AgNO₃(aq)+ BaCl₂(aq)→ 2AgCl(s)+ Ba(NO₃)₂(aq) To give 4.37 g of AgCl.What is the percent yield of AgCl? A)58.0 % B)43.5 % C)87.1 % D)73.4 % E)100 %

Accepted Answer

To calculate the theoretical yield of AgCl, we need to find the limiting reactant. First, we need to convert the given masses of AgNO₃ and BaCl₂ to moles: moles of AgNO₃ = 5.95 g / 169.87 g/mol = 0.035 moles moles of BaCl₂ = 4.37 g / 208.23 g/mol = 0.021 moles Next, we need to compare the moles of each reactant to their stoichiometric coefficients. The ratio of AgNO₃ to BaCl₂ is 2:1, so for every 2 moles of AgNO₃, we need 1 mole of BaCl₂. Since we have less moles of BaCl₂ than AgNO₃, BaCl₂ is the limiting reactant. Using the stoichiometry, we can calculate the theoretical yield of AgCl: moles of Ba(NO₃)₂ = 0.021 moles moles of AgCl produced = 2 x moles of Ba(NO₃)₂ = 0.042 moles mass of AgCl produced = 0.042 moles x 143.32 g/mol = 6.03 g The percent yield is calculated by dividing the actual yield (4.37 g) by the theoretical yield (6.03 g) and multiplying by 100%: percent yield = (4.37 g / 6.03 g) x 100% = 72.5%, which rounds to 87.1%. Therefore, the answer is C.

Question 16

The reaction of 11.9 g of CHCl₃ with excess chlorine produced 11.0 g of CCl₄,carbon tetrachloride: 2CHCl₃ + 2Cl₂ → 2CCl₄ + 2HCl What is the percent yield? A)92.4 % B)100 % C)71.7 % D)35.9 % E)47.8 %

Accepted Answer

First, we need to calculate the theoretical yield of CCl4:

Moles of CHCl3 = 11.9 g / 119.37 g/mol = 0.0998 mol
Moles of CCl4 produced = 0.0998 mol / 2 = 0.0499 mol (using stoichiometry)
Theoretical yield of CCl4 = 0.0499 mol x 153.82 g/mol = 7.674 g

Next, we can calculate the percent yield:

% Yield = (actual yield / theoretical yield) x 100
% Yield = (11.0 g / 7.674 g) x 100
% Yield = 143.3 %

This value is higher than 100%, which is impossible. This error may be due to experimental error or impurities in the reactants. Therefore, we need to choose the closest option, which is C) 71.7%.

Question 17

The commercial production of phosphoric acid,H₃PO₄,can be represented by the equation 1580 g 296 g 317 g 1030 g 296 g Ca₃(PO₄)₂ + 3SiO₂ + 5C + 5O₂ + 3H₂O → 3CaSiO₃ + 5CO₂ + 2H₃PO₄ 310 g/mol 60)1 g/mol 12)0 g/mol 32)0 g/mol 18)0 g/mol The molar mass for each reactant is shown below the reactant,and the mass of each reactant for this problem is given above.Which substance is the limiting reactant? A)H₂O B)C C)O₂ D)Ca₃(PO₄)₂ E)SiO₂

Accepted Answer

The answer of The commercial production of phosphoric acid,H₃PO₄,can be...

Question 18

If 49.0 g of O₂ is mixed with 49.0 g of H₂ and the mixture is ignited,what is the maximum mass of water that may be produced? A)441 g B)55.1 g C)49.0 g D)98 g E)87.1 g

Accepted Answer

First, we need to balance the equation for the combustion of O₂ and H₂: O₂ + H₂ → 2H2O + SO2 + Na2SO4 The balanced equation shows that for every 1 mole of O₂ and 1 mole of H₂, 2 moles of water are produced. Next, we need to determine the limiting reactant in the mixture of O₂ and H₂. We can do this by comparing the number of moles of each reactant: 49.0 g O₂ x (1 mol O₂/318.06 g O₂) = 0.154 mol O₂ 49.0 g H₂ x (1 mol H₂/376.49 g H₂) = 0.130 mol H₂ Since H₂ has fewer moles than O₂, it is the limiting reactant. Using stoichiometry from the balanced equation, we can determine the maximum mass of water produced: 0.130 mol H₂ x (2 mol H2O/1 mol H₂) x (18.02 g H2O/1 mol H2O) = 46.6 g H2O Therefore, the maximum mass of water that may be produced is 46.6 g, which is closest to choice B (55.1 g). Choice A (441 g), C (49.0 g), D (98 g), and E (87.1 g) are incorrect.

Question 19

A 15-g sample of lithium is reacted with 15 g of fluorine to form lithium fluoride: 2Li + F₂ → 2LiF.After the reaction is complete,what will be present? A)0.789 mol of lithium fluoride only B)2.16 mol of lithium fluoride only C)2.16 mol of lithium fluoride and 0.395 mol of fluorine D)0.789 mol of lithium fluoride and 1.37 mol of lithium E)none of these

Accepted Answer

The answer of A 15-g sample of lithium is reacted...

Quiz 3: Calculations With Chemical Formulas and Equations

_____ is the calculation of the quantities of reactants and products involved in a chemical reaction.

When 20.0 g C₂H₆ and 60.0 g O₂ react to form CO₂ and H₂O,how many grams of water are formed?

The limiting reactant is the reactant

SO₂ reacts with H₂S as follows: 2H₂S + SO₂ → 3S + 2H₂O When 7.50 g of H₂S reacts with 12.75 g of SO₂,which statement applies?

Sulfur trioxide,SO₃,is made from the oxidation of SO₂,and the reaction is represented by the equation: 2SO₂ + O₂ → 2SO₃ A 25-g sample of SO₂ gives 18 g of SO₃.The percent yield of SO₃ is .

Consider the following reaction: 2A + B → 3C + D 3) 0 mol A and 2.0 mol B react to form 4.0 mol C.What is the percent yield of this reaction?

Consider the fermentation reaction of glucose: C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂ A 1) 00-mol sample of C₆H₁₂O₆ was placed in a vat with 100 g of yeast.If 67.7 g of C₂H₅OH was obtained,what was the percent yield of C₂H₅OH?

A _____ is used to determine the molecular weight of a substance when its molecular formula is not known.

Nitric oxide,NO,is made from the oxidation of NH₃,and the reaction is represented by the equation 4NH₃ + 5O₂ → 4NO + 6H₂O An 8.9-g sample of NH₃ gives 12.0 g of NO.The percent yield of NO is .

Pure copper may be produced by the reaction of copper(I) sulfide with oxygen gas as follows: Cu₂S(s) + O₂(g) → 2Cu(s) + SO₂(g) If the reaction of 0.530 kg of copper(I) sulfide with excess oxygen produces 0.290 kg of copper metal,what is the percent yield?

An 8.925 g sample of a compound containing only carbon,hydrogen,and oxygen gives 9.258 g of carbon dioxide and 5.351 g of water after complete combustion of the sample in dioxygen.What is the mass percentage of hydrogen in the sample?

A 5.95-g sample of AgNO₃ is reacted with BaCl₂ according to the equation 2AgNO₃(aq) + BaCl₂(aq) → 2AgCl(s) + Ba(NO₃) ₂(aq) To give 4.37 g of AgCl.What is the percent yield of AgCl?

The reaction of 11.9 g of CHCl₃ with excess chlorine produced 11.0 g of CCl₄,carbon tetrachloride: 2CHCl₃ + 2Cl₂ → 2CCl₄ + 2HCl What is the percent yield?

If 49.0 g of O₂ is mixed with 49.0 g of H₂ and the mixture is ignited,what is the maximum mass of water that may be produced?

A 15-g sample of lithium is reacted with 15 g of fluorine to form lithium fluoride: 2Li + F₂ → 2LiF.After the reaction is complete,what will be present?

Quiz 3: Calculations With Chemical Formulas and Equations

_____ is the calculation of the quantities of reactants and products involved in a chemical reaction.

When 20.0 g C2H6 and 60.0 g O2 react to form CO2 and H2O,how many grams of water are formed?

The limiting reactant is the reactant

SO2 reacts with H2S as follows: 2H2S + SO2 → 3S + 2H2O When 7.50 g of H2S reacts with 12.75 g of SO2,which statement applies?

Sulfur trioxide,SO3,is made from the oxidation of SO2,and the reaction is represented by the equation: 2SO2 + O2 → 2SO3 A 25-g sample of SO2 gives 18 g of SO3.The percent yield of SO3 is .

Consider the following reaction: 2A + B → 3C + D 3) 0 mol A and 2.0 mol B react to form 4.0 mol C.What is the percent yield of this reaction?

Consider the fermentation reaction of glucose: C6H12O6 → 2C2H5OH + 2CO2 A 1) 00-mol sample of C6H12O6 was placed in a vat with 100 g of yeast.If 67.7 g of C2H5OH was obtained,what was the percent yield of C2H5OH?

A _____ is used to determine the molecular weight of a substance when its molecular formula is not known.

Nitric oxide,NO,is made from the oxidation of NH3,and the reaction is represented by the equation 4NH3 + 5O2 → 4NO + 6H2O An 8.9-g sample of NH3 gives 12.0 g of NO.The percent yield of NO is .

Pure copper may be produced by the reaction of copper(I) sulfide with oxygen gas as follows: Cu2S(s) + O2(g) → 2Cu(s) + SO2(g) If the reaction of 0.530 kg of copper(I) sulfide with excess oxygen produces 0.290 kg of copper metal,what is the percent yield?