Question

Gaseous butane CH3CH22CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 24. g of butane is mixed with 44.3 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Round your answer to 3 significant digits.

Answer 1

Answer:

Maximum amount of $CO_{2}$ can be produced is 37.5 g

Explanation:

Balanced equation: $2C_{4}H_{10}+13O_{2}\rightarrow 8CO_{2}+10H_{2}O$

Molar mass of butane ($C_{4}H_{10}$)  = 58.12 g/mol

Molar mass of $O_{2}$ = 32 g/mol

Molar mass of $CO_{2}$ = 44.01 g/mol

So, 24 g of butane  = $\frac{58.12}{24}mol$ of butane = 2.422 mol of butane

Also, 44.3 g of $O_{2}$  = $\frac{44.3}{32}mol$ of $O_{2}$ = 1.384 mol of $O_{2}$

According to balanced equation-

2 moles of butane produce 8 mol of $CO_{2}$

So, 2.422 moles of butane produce $(\frac{8}{2}\times 2.422)moles$ of $CO_{2}$ = 9.688 moles of $CO_{2}$

13 moles of $O_{2}$ produce 8 mol of $CO_{2}$

So, 1.384 moles of $O_{2}$ produce $(\frac{8}{13}\times 1.384)moles$ of $CO_{2}$ = 0.8517 moles of $CO_{2}$

As least number of moles of $CO_{2}$ are produced from $O_{2}$ therefore $O_{2}$ is the limiting reagent.

So, maximum amount of $CO_{2}$ can be produced = 0.8517 moles = $(44.01\times 0.8517)g=37.5 g$