Question

121.0  g of NH3 is mixed with 207.1 g of O2 and allowed to react. The reaction is complete when the NH3 is consumed.  99.56 g of N2 is formed and 36.60 g O2 remain unreacted. What mass of H2O is produced? in grams

Answer 1

The balanced equation should be,
4 NH₃ + 3 O₂ -----------> 2 N₂ + 6 H₂O
First find out the limiting reactant,
moles of NH₃ reacted = 121gm / 17.031 gm/mol = 7.1046 mol
moles of O₂ reacted = (207.1-36.60)gm / 32 gm/mol = 5.3281 mol
Therefore O₂ is the limiting reactant,
According to the reaction stoichiometry,
3 moles of O₂ gives rise to 6 moles of H₂O
So, 5.3281 moles of O₂ will give ? moles of H₂O
Cross multiply to get the answer,
(5.3281*6)/3 = 10.6562 mol of H₂O are produced
Mass of H₂O produced in grams = 10.6562mol * 18gm/mol = 191.8116 gms