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babymother [125]
3 years ago
10

Acetylene reacts with oxygen to form carbon dioxide and water according to the following unbalanced reaction:

Chemistry
1 answer:
Lelechka [254]3 years ago
7 0

Answer:

Mass of excess reactant leftover after reaction is complete = 34.2 g \mathrm{C}_{2} \mathrm{H}_{2}.

Explanation:

Acetylene reacts with oxygen to form carbon dioxide and water, equation given is unbalanced

\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})

Firstly equation is balanced:

2 \mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 4 \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})

\text { Moles of } \mathrm{CO}_{2} \text { from } \mathrm{C}_{2} \mathrm{H}_{2}

\text { Molar mass of } \mathrm{C}_{2} \mathrm{H}_{2} \text { is } 26.04 \mathrm{g} / \mathrm{mol}

\text { Moles of } \mathrm{C}_{2} \mathrm{H}_{2}=38.7 \mathrm{gm} \mathrm{C}_{2} \mathrm{H}_{2} \times \frac{1 \mathrm{mol} c_{2} \mathrm{H}_{2}}{26.04 \mathrm{gm} \mathrm{c}_{2} \mathrm{H}_{2}}=1.48 \mathrm{mol} \mathrm{C}_{2} \mathrm{H}_{2}

1.48 \mathrm{mol} \mathrm{C}_{2} \mathrm{H}_{2} \times \frac{4 \mathrm{mol} \mathrm{Co}_{2}}{2 \mathrm{mol} C_{2} \mathrm{H}_{2}}=2.96 \mathrm{mol} \mathrm{CO}_{2}

\text { Moles of } \mathrm{CO}_{2} \text { from } \mathrm{O}_{2}:

\text { Molar mass of } \mathrm{O}_{2} \text { is } 32.00 \mathrm{g} / \mathrm{mol}

\text { Moles of } \mathrm{O}_{2}=13.7 \mathrm{gm} \mathrm{O}_{2} \times \frac{1 \mathrm{mol} O_{2}}{32.00 \mathrm{gm} 0_{2}}=0.428 \mathrm{mol} \mathrm{O}_{2}

\text { Here we come to know that } 4 \mathrm{mol} \mathrm{CO}_{2} \equiv 5 \mathrm{mol} \mathrm{O}_{2}, \text { hence }

0.428 \mathrm{mol} \mathrm{O}_{2} \times \frac{4 \mathrm{mol} \mathrm{co}_{2}}{5 \mathrm{mol} O_{3}}\left(4 \mathrm{mol} \mathrm{CO}_{2}\right) /\left(5 \mathrm{mol} \mathrm{O}_{2}\right)=0.3424 \mathrm{mol} \mathrm{CO}_{2}

Here we come to know O2 is limiting reactant as it gives smaller amount of \mathrm{CO}_{2} So \mathrm{C}_{2} \mathrm{H}_{2} is excess reactant.

Mass of excess reactant utilized:

0.428 \mathrm{mol} \mathrm{O}_{2} \times \frac{2 \mathrm{mol} C_{2} \mathrm{H}_{2}}{5 \mathrm{mol} O_{2}}=0.1712 \mathrm{mol} \mathrm{C}_{2} \mathrm{H}_{2}

0.1712 \mathrm{mol} \mathrm{C}_{2} \mathrm{H}_{2} \times \frac{26.04 \mathrm{gm} \mathrm{c}_{2} \mathrm{H}_{2}}{1 \mathrm{mol} C_{2} \mathrm{H}_{2}}=4.45 \mathrm{gm} \mathrm{C}_{2} \mathrm{H}_{2}

Mass of excess reactant leftover after reaction is complete:

38.7 \mathrm{gm} \mathrm{C}_{2} \mathrm{H}_{2}-4.45 \mathrm{gm} \mathrm{C}_{2} \mathrm{H}_{2}=34.2 \mathrm{gm} \mathrm{C}_{2} \mathrm{H}_{2}.

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If 24.1 g of sodium hydroxide react with 22.0 g of hydrochloric acid to form 35.3g
disa [49]

Answer:

10.85 g of H2O.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

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Next, we shall determine the mass of NaOH that reacted and the mass of H2O produced from the balanced equation.

These can be obtained as illustrated below:

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Summary:

From the balanced equation above,

40 g of NaOH reacted to produce 18 g of H2O.

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From the balanced equation above,

40 g of NaOH reacted to produce 18 g of H2O.

Therefore, 24.1 g of NaOH will react to produce = (24.1 × 18)/40 = 10.85 g of H2O.

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Explanation:

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2 years ago
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