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Olin [163]
3 years ago
10

The reaction of 5.40 g of carbon with excess O2 yields 13.6 g of CO2. What is the percent yield of this reaction?

Chemistry
1 answer:
vazorg [7]3 years ago
4 0
C + O2= CO2
n  =  \frac{m}{mw}
n =  \frac{5.4}{12}  \\ n  = 0.45 \: mol \: of \: carbon
n =  \frac{13.6}{12 + 16 \times 2} \\ n =  \frac{13.6}{44}  \\ n = 0.31 \: mol \: of \: carbon \: dioxide
CO2 is limit
5.4-3.72= 1.68 g of C is excess
5.4 g = 100%
3.72 g = x
x=68.9 %
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Sindrei [870]

Answer:

The percent yield of the reaction is 35 %

Explanation:

In the reaction, 1 mol of hydrazine reacts with 1 mol O₂ to produce 1 mol of nitrogen and 2 moles of water.

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2.05 g . 1mol/ 32 g = 0.0640 mol

In the 100% yield, 1 mol of hydrazine produce 1 mol of N₂ so If I used 0.0640 moles of reactant, I made 0.0640 moles of products.

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3 years ago
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Let x be the number of gold atoms. Plug in the values to a calculator.

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3 years ago
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Amanda [17]

Answer:

= 0.014 g of BaCO3

Explanation:

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But;

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Therefore;

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Thus;

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Therefore;

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3 years ago
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