Answer:
126.4 g of [tex]C_{4}H_{10}[/tex] are required
Explanation:
Balanced reaction: [tex]2C_{4}H_{10}+13O_{2}\rightarrow 8CO_{2}+10H_{2}O[/tex]
According to balanced reaction-
8 moles of [tex]CO_{2}[/tex] are produced from 2 moles of [tex]C_{4}H_{10}[/tex]
So, 8.70 moles of [tex]CO_{2}[/tex] are produced from [tex](\frac{2}{8}\times 8.70)[/tex] moles of [tex]C_{4}H_{10}[/tex] or 2.175 moles of [tex]C_{4}H_{10}[/tex]
Molar mass of [tex]C_{4}H_{10}[/tex] = 58.12 g/mol
So, mass of [tex]C_{4}H_{10}[/tex] required = [tex](2.175\times 58.12)g[/tex] = 126.4 g
Hence 126.4 g of [tex]C_{4}H_{10}[/tex] are required
