Question

Determine the number of grams of C4H10 that are required to completely react to produce 8.70 mol of CO2 according to the following combustion reaction: 2C4H10 13O2 --> 8CO2 10H2O

Answer 1

Answer:

126.4 g of $C_{4}H_{10}$ are required

Explanation:

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

According to balanced reaction-

8 moles of $CO_{2}$ are produced from 2 moles of $C_{4}H_{10}$

So, 8.70 moles of $CO_{2}$ are produced from $(\frac{2}{8}\times 8.70)$ moles of $C_{4}H_{10}$ or 2.175 moles of $C_{4}H_{10}$

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

So, mass of $C_{4}H_{10}$ required = $(2.175\times 58.12)g$ = 126.4 g

Hence 126.4 g of $C_{4}H_{10}$ are required