propane (c3h8), a common fuel, reacts with oxygen to form carbon dioxide and water according to the equation…

propane (c3h8), a common fuel, reacts with oxygen to form carbon dioxide and water according to the equation below. c3h8 + 5o2 → 3co2 + 4h2o how many grams of oxygen are required to produce 37.15 g co2? 37.15 g co2 = g o2 done

propane (c3h8), a common fuel, reacts with oxygen to form carbon dioxide and water according to the equation below. c3h8 + 5o2 → 3co2 + 4h2o how many grams of oxygen are required to produce 37.15 g co2? 37.15 g co2 = g o2 done

Answer

Explanation:

Step1: Calculate moles of CO₂

The molar - mass of CO₂ is (M_{CO_2}=(12 + 2\times16)\ g/mol=44\ g/mol). The number of moles of CO₂, (n_{CO_2}=\frac{m_{CO_2}}{M_{CO_2}}), where (m_{CO_2} = 37.15\ g). So (n_{CO_2}=\frac{37.15\ g}{44\ g/mol}=0.8443\ mol).

Step2: Determine mole - ratio

From the balanced chemical equation (C_3H_8 + 5O_2\rightarrow3CO_2+4H_2O), the mole - ratio of (O_2) to (CO_2) is (\frac{n_{O_2}}{n_{CO_2}}=\frac{5}{3}).

Step3: Calculate moles of O₂

(n_{O_2}=\frac{5}{3}n_{CO_2}). Substituting (n_{CO_2}=0.8443\ mol), we get (n_{O_2}=\frac{5}{3}\times0.8443\ mol = 1.4072\ mol).

Step4: Calculate mass of O₂

The molar - mass of O₂ is (M_{O_2}=2\times16\ g/mol = 32\ g/mol). The mass of O₂, (m_{O_2}=n_{O_2}\times M_{O_2}). So (m_{O_2}=1.4072\ mol\times32\ g/mol = 45.03\ g).

Answer:

45.03