Question

2 H2(g) + O2(g) → 2 H2O(g)

Answer 1

Answer:

77 L of water can be made.

Explanation:

Molar mass of $O_{2}$ = 32 g/mol

So, 55 g of $O_{2}$ = $\frac{55}{32}$ mol of  $O_{2}$ = 1.72 mol of  $O_{2}$

As hydrogen is present in excess amount therefore  $O_{2}$ is the limiting reagent.

According to balanced equation, 1 mol of  $O_{2}$ produces 2 mol of $H_{2}O$.

So, 1.72 mol of $O_{2}$ produce $(2\times 1.72)$ mol of $H_{2}O$ or 3.44 mol of $H_{2}O$.

Let's assume $H_{2}O$ gas behaves ideally at STP.

Then, $P_{H_{2}O}.V_{H_{2}O}=n_{H_{2}O}.R.T$   , where P, V, n, R and T represents pressure, volume, no. of moles, gas constant and temperature in kelvin scale respectively.

At STP, pressure is 1 atm and T is 273 K.

Here, $n_{H_{2}O}$ = 3.44 mol and R = 0.0821 L.atm/(mol.K)

So, $(1atm)\times V_{H_{2}O}=(3.44mol)\times (0.0821L.atm.mol^{-1}.K^{-1})\times (273K)$

  $\Rightarrow$  $V_{H_{2}O}=77L$

Option (b) is correct.