Question

consider the formation of liquid water through the following reaction: 2 h₂(g) + o₂(g) → 2 h₂o(l) if the reaction occurs at 85°c and 1.5 atm of pressure, what volume of hydrogen gas is needed to create 50.0 g of water? r = 0.821 (\frac{l cdot atm}{mol cdot k}) (\bigcirc) 13 l (\bigcirc) 115 l (\bigcirc) 54 l (\bigcirc) 2.8 l

Explanation:

Step1: Find moles of \(H_2O\)

Molar mass of \(H_2O\) is \(18.02\ g/mol\). Moles of \(H_2O = \frac{50.0\ g}{18.02\ g/mol} \approx 2.775\ mol\).

Step2: Relate moles of \(H_2\) and \(H_2O\)

From reaction \(2H_2 + O_2
ightarrow 2H_2O\), moles of \(H_2\) = moles of \(H_2O\) = \(2.775\ mol\).

Step3: Convert temperature to Kelvin

\(T = 85^\circ C + 273.15 = 358.15\ K\).

Step4: Use ideal gas law \(PV = nRT\)

Solve for \(V\): \(V = \frac{nRT}{P}\). Substitute \(n = 2.775\ mol\), \(R = 0.821\frac{L\cdot atm}{mol\cdot K}\), \(T = 358.15\ K\), \(P = 1.5\ atm\).
\(V = \frac{2.775\ mol \times 0.821\frac{L\cdot atm}{mol\cdot K} \times 358.15\ K}{1.5\ atm}\)
Calculate numerator: \(2.775 \times 0.821 \times 358.15 \approx 2.775 \times 294.0 \approx 816.85\)
Divide by \(1.5\): \(V \approx \frac{816.85}{1.5} \approx 544\)? Wait, no, wait. Wait, moles of \(H_2\): Wait, reaction is \(2H_2\) gives \(2H_2O\), so moles of \(H_2\) = moles of \(H_2O\)? Wait no, 2 moles \(H_2\) produce 2 moles \(H_2O\), so 1:1. Wait but my calculation gave wrong. Wait, let's recalculate moles of \(H_2O\): \(50 / 18.02 ≈ 2.775\) moles. Then moles of \(H_2\) is also 2.775? Wait no, 2 moles \(H_2\) make 2 moles \(H_2O\), so yes. Then \(V = (2.775 0.821 358.15)/1.5\). Let's compute 0.821358.15 ≈ 0.821358 ≈ 294. So 2.775294 ≈ 816, divided by 1.5 ≈ 544? But options are 13, 115, 54, 2.8. Wait, I must have messed up moles. Wait, wait, 50g of water: moles is 50/18 ≈ 2.77, but maybe the reaction is 2 moles \(H_2\) make 2 moles \(H_2O\), so moles \(H_2\) = moles \(H_2O\). But maybe I made a mistake in moles. Wait, no, let's check again. Wait, ideal gas law: \(V = nRT/P\). Let's plug numbers: n=2.775, R=0.821, T=358, P=1.5. So 2.7750.821=2.28, 2.28*358=816, 816/1.5=544. But options have 54L. Wait, maybe I messed up moles. Wait, maybe the question is to make 50g water, but maybe the moles of \(H_2\) is half? No, reaction is 2H2 + O2 -> 2H2O, so 2 moles H2 produce 2 moles H2O, so 1:1. Wait, maybe the mass is 5.0g? No, the question says 50.0g. Wait, maybe the answer is 54L, which is close to 544? No, that's a factor of 10. Wait, maybe I miscalculated moles. 50g / 18g/mol ≈ 2.775 moles. Wait, maybe the temperature is 85K? No, 85C is 358K. Wait, maybe R is 0.0821? Oh! Wait, R is 0.0821, not 0.821. Oh! That's the mistake. R is \(0.0821\frac{L\cdot atm}{mol\cdot K}\), not 0.821. That's the error. So let's correct:

\(V = \frac{nRT}{P}\), \(n = 2.775\ mol\), \(R = 0.0821\), \(T = 358\ K\), \(P = 1.5\ atm\)

\(V = \frac{2.775 \times 0.0821 \times 358}{1.5}\)

Calculate 0.0821358 ≈ 29.4, 2.77529.4 ≈ 81.6, 81.6 / 1.5 ≈ 54.4 L, which is ~54L. Ah, there we go. I used R as 0.821 instead of 0.0821. So the correct volume is approximately 54L.

Answer:

54L (the option with 54L)