Question

Use the following equation to determine how many grams of O2 are required to react with 22.50g of C7H16?

C7H16 + 11O2 --> 7CO2 + 8H2O

Answer 1

79.2 grams of oxygen gas are required to react with 22.50 grams of n-heptane.

Step-by-step explanation:

`C_7H_(16) + 11O_2\rightarrow 7CO_2 + 8H_2O`

Mass of n-heptane = 22.50 g

Moles of n-heptane =
`(22.50 g)/(100 g/mol)=0.2250 mol`

According to reaction 1 mole of n-heptane reacts with 11 moles of oxygen gas. then 0.2250 moles of n-heptane will react with:

`(11)/(1)* 0.2250 mol=2.475 mol` of oxygen gas

Mass of 2.475 moles of oxygen gas= 2.475 mol × 32 g/mol= 79.2 g

79.2 grams of oxygen gas are required to react with 22.50 grams of n-heptane.