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{lcdot atm}{molcdot k})
options:
13 l
115 l
54 l
2.8 l

Explanation:

Step1: Calculate moles of water

Molar mass of \(H_2O\) is \(18.02\space g/mol\). Moles of \(H_2O=\frac{50.0\space g}{18.02\space g/mol}\approx2.775\space mol\).

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

From reaction \(2H_2 + O_2
ightarrow2H_2O\), moles of \(H_2\) = moles of \(H_2O\) = \(2.775\space mol\).

Step3: Convert temperature to Kelvin

\(T = 85^\circ C+ 273.15 = 358.15\space K\), \(P = 1.5\space atm\), \(R = 0.821\frac{L\cdot atm}{mol\cdot K}\).

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

Solve for \(V\): \(V=\frac{nRT}{P}=\frac{2.775\space mol\times0.821\frac{L\cdot atm}{mol\cdot K}\times358.15\space K}{1.5\space atm}\).
Calculate numerator: \(2.775\times0.821\times358.15\approx2.775\times294.0\approx816.8\).
Then \(V=\frac{816.8}{1.5}\approx544\)? Wait, no, wait moles of \(H_2\): Wait reaction is \(2H_2 + O_2
ightarrow2H_2O\), so moles of \(H_2\) = moles of \(H_2O\). Wait 50g \(H_2O\) is \(\frac{50}{18}\approx2.778\) mol \(H_2O\), so moles of \(H_2\) is 2.778 mol. Then \(V=\frac{nRT}{P}=\frac{2.778\times0.821\times358.15}{1.5}\). Let's recalculate: \(0.821\times358.15\approx294\), \(2.778\times294\approx816\), \(816\div1.5 = 544\)? No, options have 54L. Wait, maybe I messed moles. Wait, reaction: 2 moles \(H_2\) produce 2 moles \(H_2O\), so 1:1. Wait 50g \(H_2O\) is \(\frac{50}{18}\approx2.778\) mol. Then \(V=\frac{nRT}{P}=\frac{2.778\times0.821\times358.15}{1.5}\). Wait 0.821358.15 = 0.821358 = approx 0.821300=246.3, 0.82158=47.6, total 293.9. Then 2.778293.9≈2.778294≈816.7. 816.7/1.5≈544? No, options are 13,115,54,2.8. Wait, maybe I made a mistake in moles. Wait, maybe the question is about hydrogen to make 50g water, but maybe I miscalculated moles. Wait 50g water is 50/18≈2.778 mol. But ideal gas law: \(V = nRT/P\). Wait R is 0.821, T is 358K, P is 1.5 atm. So \(V=(2.778 0.821 358)/1.5\). Let's compute 2.7780.821=2.281, 2.281358=816.6, 816.6/1.5=544? No, that's not matching options. Wait, maybe the reaction is 2H2 + O2 -> 2H2O, so moles of H2 is equal to moles of H2O. Wait, maybe the mass of water is 50g, but maybe I messed up the R value? Wait R is 0.821 L·atm/(mol·K). Wait, maybe the question has a typo, or I misread. Wait the options are 13L, 115L, 54L, 2.8L. Wait, maybe I miscalculated moles. Wait 50g water is 50/18≈2.778 mol. But maybe the question is about hydrogen to make 50g water, but maybe the moles of H2 is 2.778, but let's check the calculation again. Wait, 0.821 358 = 294, 294 2.778 = 816, 816 / 1.5 = 544. No, that's not matching. Wait, maybe the moles of H2 is half? No, reaction is 2H2 -> 2H2O, so 1:1. Wait, maybe the mass of water is 5.0g? Then 5/18≈0.2778 mol. Then V=(0.27780.821358)/1.5. 0.27780.821=0.228, 0.228358=81.6, 81.6/1.5=54.4L≈54L. Ah! Maybe the question is 5.0g water, not 50g. That must be a typo. So assuming 5.0g water: moles of H2O = 5/18≈0.2778 mol, moles of H2=0.2778 mol. Then V=(0.27780.821358)/1.5. 0.27780.821=0.228, 0.228358=81.6, 81.6/1.5=54.4L≈54L. So the answer is 54L.

Answer:

54L (corresponding to the option "54 L")