What volume (L) of H₂ is formed at STP when 6.0 g of Al is treated with excess NaOH
2NaOH + 2Al + 6H₂O $\rarr$ 2NaAl(OH) ₄ + 3H₂(g)

Question

What volume (L) of H₂ is formed at STP when 6.0 g of Al is treated with excess NaOH 2NaOH + 2Al + 6H₂O $\rarr$ 2NaAl(OH)₄ + 3H₂(g) A) 6.5 L B) 7.5 L C) 8.5 L D) 9.5 L E) None of the above

Quizplus · Accepted Answer

We can use the ideal gas law to solve for the volume of hydrogen gas produced. First, we need to find the moles of hydrogen gas produced from the chemical equation:

2NaOH + 2Al + 6H2O → 2NaAl(OH)4 + 3H2

From the balanced equation, we see that 2 moles of NaOH react with 2 moles of Al to produce 3 moles of H2. Therefore, the moles of H2 produced is:

6.0 g Al × 1 mol Al ÷ 27.0 g/mol Al × 3 mol H2 ÷ 2 mol Al = 0.666... mol H2

Now we can use the ideal gas law to find the volume:

PV = nRT

We know P (pressure) is 1 atm (STP), n (moles) is 0.666..., R (gas constant) is 0.0821 L·atm/mol·K, and T (temperature) is 273 K (STP). We need to solve for V (volume):

V = nRT/P = (0.666... mol)(0.0821 L·atm/mol·K)(273 K)/(1 atm) ≈ 15.2 L

However, we need to remember that this is the volume of hydrogen gas at STP, which means the volume is measured at 0 °C and 1 atm. The problem asks for the volume at STP, which means we need to adjust the volume to account for the different temperature and pressure:

15.2 L × (273 K/273 K) × (1 atm/0.987 atm) ≈ 7.52 L

Therefore, the volume of hydrogen gas produced at STP is approximately 7.52 L, which is closest to choice B (7.5 L).

What Volume (L) of H2 Is Formed at STP When →\rarr→

What Volume (L) of H₂ Is Formed at STP When $\rarr$