Using the balanced chemical equation: 2C_(4)H_(10)+13O_(2)arrow 10H_(2)O+8CO_(2) Calculate the number of grams of C_(4)H_(10) needed to produce 5.12 moles of CO_(2) 58.0gC_(4)H_(10) 2.56gC_(4)H_(10) 74.2gC_(4)H_(10) 2gC_(4)H_(10)

To solve this problem, we need to use the balanced chemical equation and the given information to calculate the number of grams of $C_{4}H_{10}$ needed to produce 512 moles of $CO_{2}$.<br /><br />Given information:<br />- Balanced chemical equation: $2C_{4}H_{10} + 13O_{2} \rightarrow 10H_{2}O + 8CO_{2}$<br />- Number of moles of $CO_{2}$ produced: 512 moles<br /><br />Step 1: Determine the mole ratio between $C_{4}H_{10}$ and $CO_{2}$.<br />From the balanced chemical equation, we can see that 2 moles of $C_{4}H_{10}$ produce 8 moles of $CO_{2}$. Therefore, the mole ratio of $C_{4}H_{10}$ to $CO_{2}$ is 2:8, or 1:4.<br /><br />Step 2: Calculate the number of moles of $C_{4}H_{10}$ needed to produce 512 moles of $CO_{2}$.<br />Since the mole ratio of $C_{4}H_{10}$ to $CO_{2}$ is 1:4, we can calculate the number of moles of $C_{4}H_{10}$ needed as follows:<br />Number of moles of $C_{4}H_{10}$ = Number of moles of $CO_{2}$ / 4<br />Number of moles of $C_{4}H_{10}$ = 512 moles / 4 = 128 moles<br /><br />Step 3: Calculate the mass of $C_{4}H_{10}$ needed to produce 512 moles of $CO_{2}$.<br />The molar mass of $C_{4}H_{10}$ is 58.0 g/mol.<br />Mass of $C_{4}H_{10}$ = Number of moles of $C_{4}H_{10}$ × Molar mass of $C_{4}H_{10}$<br />Mass of $C_{4}H_{10}$ = 128 moles × 58.0 g/mol = 7,424 g<br /><br />Therefore, the correct answer is:<br />$74.2gC_{4}H_{10}$