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How can Cold air be made to rise

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Answer:

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2) Limiting reagent: $NO$

3) Ammount of excess reagent: $m_{N2}=4.274 g$

Explanation:

The reaction

$2 NO (g) + 2 H_2 (g) \longrightarrow N_2 (g) + 2 H_2O (g)$

Moles of nitrogen monoxide

Molecular weight: $M_{NO}=30 g/mol$

$n_{NO}=\frac{m_{NO}}{M_{NO}}$

$n_{NO}=\frac{22.9 g}{30 g/mol}=0.763 mol$

Moles of hydrogen

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$n_{H2}=\frac{m_{H2}}{M_{H2}}$

$n_{H2}=\frac{.5.8 g}{2 g/mol}=2.9 mol$

Mol rate of H2 and NO is 1:1 => hydrogen gas is in excess

1) Maximun ammount of nitrogen gas => when all NO reacted

$m_{N2}=0.763 mol NO* \frac{1 mol N_2}{2 mol NO}*\frac{28 g N_2}{mol N_2}$

$m_{N2}=10.682 g N_2$

2) Limiting reagent: $NO$

3) Ammount of excess reagent:

$m_{N2}=(2.9 mol - 0.763 mol NO* \frac{1 mol H_2}{1 mol NO})*\frac{2 g H_2}{mol H_2}$

$m_{N2}=4.274 g$

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