Calculate the volume of H₂(g) at 273 K and 2.00 atm that will be formed when 275 mL of 0.725 M HCl solution reacts with excess Mg to give hydrogen gas and aqueous magnesium chloride. A) 0.56 L B) 1.12 L C) 2.23 L D) 3.54 L E) 4.47 L

Question

Quizplus · Accepted Answer

First, we need to balance the equation: 
2HCl(aq) + Mg(s) → H2(g) + MgCl2(aq)

Next, we need to find the number of moles of HCl used: 
0.725 mol/L x 0.275 L = 0.199375 moles

According to stoichiometry, 2 moles of HCl produces 1 mole of H2, which means 0.199375 moles of HCl will produce 0.0996875 moles of H2.

We can use the ideal gas law to calculate the volume of H2 gas produced: 
PV = nRT 
V = nRT/P 
V = (0.0996875 mol)(0.08206 L·atm/mol·K)(273 K)/(2.00 atm) 
V = 1.12 L

Therefore, the correct answer is B) 1.12 L.

Calculate the Volume of H2(g) at 273 K and 2

Calculate the Volume of H₂(g) at 273 K and 2