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sergey [27]
3 years ago
9

Balance the following redox reaction in acidic solution: H+(aq)+Zn(s)→H2(g)+Zn2+(aq)

Chemistry
1 answer:
erma4kov [3.2K]3 years ago
5 0
2H⁺(aq) + Zn(s) ----> H2(g) + Zn²⁺(aq)

2H⁺(aq) +2e⁻ ----->H2(g)            reduction
<span><u>Zn(s)            ------>Zn²⁺ +2e⁻      oxidation</u>
</span>2H⁺(aq) + Zn(s) -----> H2(g) + Zn²⁺
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For the following reaction, 22.9 grams of nitrogen monoxide are allowed to react with 5.80 grams of hydrogen gas. nitrogen monox
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Answer:

1) Maximun ammount of nitrogen gas: m_{N2}=10.682 g N_2

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<u>The reaction </u>

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

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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) <u>Maximun ammount of nitrogen gas</u> => 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) <u>Limiting reagent</u>: NO

3) <u>Ammount of excess reagent</u>:

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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