Respuesta :
Answer:
52.36 grams of water was produced when 32 grams of propane was burned in excess of oxygen gas.
Explanation:
[tex]C_3H_8+5O_2\rightarrow 3CO_2+4H_2O[/tex]
Moles of propane = [tex]\frac{32 g}{44 g/mol}=0.7273 mol[/tex]
According to reaction, 1 mole of propane gives 4 moles of water.
Then 0.7273 moles of propane will give:
[tex]\frac{4}{1}\times 0.7273 mol=2.909 mol[/tex] 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.
Answer:
There will be produced 52.29 grams of water
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
