Question

32 grams of propane (C3H8) is burned in excess oxygen gas to produce how many grams of water?

Answer 1

There will be produced 52.29 grams of water

Step-by-step explanation:

Step 1: Data given

Mass of propane = 32.00 grams

Molar mass of propane = 44.1 g/mol

Oxygen = in excess

Step 2: The balanced equation

C3H8 + 5O2 → 3CO2 + 4H2O

Step 3: Calculate moles of propane

Moles propane = Mass propane / molar mass propane

Moles propane = 32.00 grams / 44.10 g/mol

Moles propane = 0.7256 moles

Step 4: Calculate moles of water

The limiting reactant is propane.

For 1 mol of propane consumed, we need 5 moles of O2 to produce 3 moles of CO2 and 4 moles of H2O

For 0.7256 moles of propane consumed we'll have 4* 0.7256 = 2.902

Step 5: Calculate mass of H2O

Mass H2O = moles H2O * molar mass H2O

Mass H2O = 2.902 moles * 18.02 g/mol

Mass H2O = 52.29 grams

There will be produced 52.29 grams of water

Answer 2

52.36 grams of water was produced when 32 grams of propane was burned in excess of oxygen gas.

Step-by-step explanation:

`C_3H_8+5O_2\rightarrow 3CO_2+4H_2O`

Moles of propane =
`(32 g)/(44 g/mol)=0.7273 mol`

According to reaction, 1 mole of propane gives 4 moles of water.

Then 0.7273 moles of propane will give:

`(4)/(1)* 0.7273 mol=2.909 mol` of water

Mass of 2.909 moles of water:

= 2.909 mol × 18 g/mol = 52.36 g

52.36 grams of water was produced when 32 grams of propane was burned in excess of oxygen gas.