Question

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

Answer 1

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

Answer 2

Answer: 4 moles of
`CO_2` are produced.

Step-by-step explanation:

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}=(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 =
`(8)/(2)* 1=4moles` of carbon dioxide

Hence, 4 moles of
`CO_2` are produced.