Question

For the following reaction, 9.60 grams of butane (C4H10) are allowed to react with 17.0 grams of oxygen gas. butane (C4H10) (g) + oxygen (g) carbon dioxide (g) + water (g) What is the maximum amount of carbon dioxide that can be formed? grams

Answer 1

Answer:

14.4 g of $CO_{2}$ can be produced.

Explanation:

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

                                        Molar mass (g/mol)

                  $C_{4}H_{10}$                    58.12

                    $O_{2}$                         32

                  $CO_{2}$                       44.01

So, 9.60 g of $C_{4}H_{10}$ = $\frac{9.60}{58.12}mol=0.165mol$

      17.0 g of $O_{2}$ = $\frac{17.0}{32}mol=0.531mol$

According to balanced equation-

2 moles of $C_{4}H_{10}$ produce 8 moles of $CO_{2}$

So, 0.165 moles of $C_{4}H_{10}$ produce $(\frac{8}{2}\times 0.165)mol$ of $CO_{2}$  or 0.660 moles of $CO_{2}$

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

So, 0.531 moles of $O_{2}$ produce $(\frac{8}{13}\times 0.531)moles$ of $CO_{2}$ or 0.327 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 $CO_{2}$ that can be formed = 0.327 moles

                                                                              = $(0.327\times 44.01)g$

                                                                              = 14.4 g