Question

How 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)?

a) 2c6h14
b) 19o2
c) 12co2
d) 14h2o

Answer 1

Approximately 441.32 grams of CO2 are produced in the combustion of 72.0 grams of C6H14.

Step-by-step explanation:

To determine the amount of CO2 produced in the combustion of C6H14, we need to use the given molar masses and balanced chemical equation. From the equation, we can see that for every 1 mol of C6H14, 12 mol of CO2 are produced. We can use this ratio to calculate the amount of CO2 produced in grams:

1. Calculate the number of moles of C6H14 by dividing the given mass by its molar mass: 72.0 g / 86.2 g/mol = 0.836 mol.
2. Use the mole ratio from the balanced chemical equation to find the number of moles of CO2 produced: 0.836 mol C6H14 * (12 mol CO2 / 1 mol C6H14) = 10.03 mol CO2.
3. Convert the number of moles of CO2 to grams using its molar mass: 10.03 mol CO2 * 44.0 g/mol = 441.32 g CO2.

Therefore, approximately 441.32 grams of CO2 are produced in the combustion of 72.0 grams of C6H14.