Please answer both questions Thankyou

The limiting reactant is oxygen gas and the reaction would produce 3.932 x 10 ^ 7 moles of water

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