Answer:
The limiting reactant is oxygen gas and the reaction would produce 3.932 x 10 ^ 7 moles of water
Step-by-step explanation:
Step 1: Convert everything into moles
nH2(l) = 1.06 x 10^8 g / 2.016 g/mol = 5.26 x 10^7 mols
nO2(l) = 6.29 x 10^8 g / 32.00 g/ mol = 1.966 x 10^7 mols
Step 2: Find the limiting reagent
The limiting reagent would be oxygen gas from
the balanced equation because we have less moles of oxygen gas needed to fully combust with the hydrant gas
Step 3: Stoichiometry time
The mole ratio from oxygen gas to water is 1:2
This means that for every mole of oxygen gas two moles of water is produced
We need to multiply the moles of oxygen gas by two to find out how many moles of water has been produced
nH2O = nO2 x 2
nH2O = 1.966x10^7 x 2
nH2O = 3.932x10^7
Step 4: Therefore statement
Therefore the limiting reactant is oxygen gas and the reaction would produce 3.932 x 10 ^ 7 moles of water