Which of the following statements is/are correct?A mixture containing 64.0 g H2 and 64.0 g O2 is ignited so that water is formed as follows:2H2+O2→ 2H2O

nH2=642=32 molnO2=6432= 2mol1 mol of O2 requires = 2 mol of H22 mol of O2 requires = 4 mol g H2Since H2 is present in excess, therefore, O2 is the limiting reagent.1. Wrong.2. So, O2 is the limiting reagent3. 2 mol of O2=4 moles of H2O = 4×18=72 g H2O moles of H2 left = 32-4=28 mol =28×2=56 g H2 Mixture contains = (72g H2O + 56 g H2) 4. Wrong

Which of the following statements is/are correct?A mixture containing 64.0 g H2 and 64.0 g O2 is ignited so that water is formed as follows:2H2+O2→ 2H2O

nH2=642=32 molnO2=6432= 2mol1 mol of O2 requires = 2 mol of H22 mol of O2 requires = 4 mol g H2Since H2 is present in excess, therefore, O2 is the limiting reagent.1. Wrong.2. So, O2 is the limiting reagent3. 2 mol of O2=4 moles of H2O = 4×18=72 g H2O moles of H2 left = 32-4=28 mol =28×2=56 g H2 Mixture contains = (72g H2O + 56 g H2) 4. Wrong

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Which of the following statements is/are correct?
A mixture containing 64.0 g H₂and 64.0 g O₂is ignited so that water is formed as follows:
$2 H_{2} + O_{2} \overset{}{\to} 2 H_{2} O$

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O₂ is the limiting reagent

The reaction mixture contains 72.0 g of H₂O and 56.0 g of unreacted H₂.

H₂ is the limiting reagent

The reaction mixture contains 56.0 g of H₂O and 72.0 g of unreacted H₂.

answer is B, C.

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Detailed Solution

$\begin{aligned} n_{H_{2}} = \frac{64}{2} = 32 mol \\ n_{O_{2}} = \frac{64}{32} = 2 mol \end{aligned}$

1 mol of O₂ requires = 2 mol of H₂

2 mol of O₂ requires = 4 mol g H₂

Since H₂ is present in excess, therefore, O₂ is the limiting reagent.

1. Wrong.

2. So, O₂ is the limiting reagent

$3 . 2 mol of O_{2} = 4 moles of H_{2} O = 4 \times 18 = 72 g H_{2} O moles of H_{2} left = 32 - 4 = 28 mol = 28 \times 2 = 56 g H_{2} Mixture contains = (72 g H_{2} O + 56 g H_{2}) 4 . Wrong$