Question

ow many grams of co2 (mm = 44.0 g/mol) are produced in the combustion of 72.0 g of c6h14 (mm = 86.2 g/mol)? 2c6h14 19o2 12co2 14h2o

Answer 1

In the combustion of C6H14, 528.108 grams of CO2 are produced from 72.0 grams of C6H14.

Step-by-step explanation:

In the combustion of C6H14, the balanced chemical equation is 2C6H14 + 19O2 → 12CO2 + 14H2O. From this equation, we can determine the molar ratio between C6H14 and CO2, which is 2 mol C6H14 : 12 mol CO2. We also know the molar masses of CO2 (44.009 g/mol) and C6H14 (86.2 g/mol). To find the mass of CO2 produced, we can use the following steps:

1. Convert the mass of C6H14 to moles using its molar mass.
2. Use the molar ratio from the balanced chemical equation to find the moles of CO2 produced.
3. Convert the moles of CO2 to mass using the molar mass of CO2.

Let's calculate it:

1. Mass of C6H14 = 72.0 g
2. Moles of C6H14 = 72.0 g / 86.2 g/mol = 0.836 mol
3. Moles of CO2 = 2 mol C6H14 * (12 mol CO2 / 2 mol C6H14) = 12 mol CO2
4. Mass of CO2 = 12 mol CO2 * 44.009 g/mol = 528.108 g

Therefore, 528.108 grams of CO2 are produced in the combustion of 72.0 grams of C6H14.