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maksim [4K]
3 years ago
9

how many moles of CO2 form when 58.0 g of butane, C4H10, burn in oxygen? 2C4H10+13O2--->8CO2+10H2O

Chemistry
2 answers:
igomit [66]3 years ago
5 0

<u>Answer: </u>4 moles of CO_2 are produced.

<u>Explanation:</u>

To calculate the number of moles, we use the formula:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

For butane:

Given mass of butane = 58g

Molar mass of butane = 58 g/mol

Putting values in above equation, we get:

\text{Moles of butane}=\frac{58g}{58g/mol}=1mole

For the given chemical equation:

2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O

By Stoichiometry of the reaction:

2 moles of butane produces 8 moles of carbon dioxide

So, 1 mole of butane will produce = \frac{8}{2}\times 1=4moles of carbon dioxide

Hence, 4 moles of CO_2 are produced.

Finger [1]3 years ago
3 0
2C4H10 + 13O2 ---> 8CO2 + 10H2O 1 mole of C4H10 = 58g According to the reaction: 2*58g of C4H10 ------------- 8 molesof CO2 58g of C4H10 ----------------- x moles of CO2 x = 4 miles od CO2
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