Question

1.What mass of oxygen is necessary to burn 106.9g of propane in a balanced fashion?

2.What is the theoretical yield of water, in grams?

3.How many grams of oxygen does it take to burn 15.4 g of glucose

Answer 1

C3H8 + 5O2 --> 4H2O + 3CO2

1 mol of fuel per 5 mol of O2

MW propane = 44

mol = mass/MW = 106.9/44 = 2.43 mol of propane

then 2.43*5 = 12.15 mol of oxygen is required

mass = mol*MW = 12.15*35 = 388.8 g of O2

2)

theoretical yield of water is

1 mol of fuel --> 4 mol of water

2.43 *4 = 9.72 mol of water are formed

mass = mol*MW = 9.72*18 = 174.96 g of water

3)

MW of glucose = 180

mol = mass/MW = 15.4/180 = 0.08555 mol of glucose

1 mol of glucose requires 6 mol of O"

then

0.08555*6 = 0.5133 mol of O2 requried

mass = mol*MW = 0.5133*32 = 16.4256 g of O2