2H2(g)+ O2(g)→ 2H2O(l) Thus, H2 is the limiting reagent

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2H2(g)+ O2(g)→ 2H2O(l) Thus, H2 is the limiting reagent 2 mol H2 = 2.44 mol H2 1.22 mol O2 x 1 mol O2 Given: 2.05 mol H2 2 mol H2O 18.015 g H2O 2.05 mol H2 x x = 36.93 g H2O 2 mol H2 1 mol H2O Thus, H2 is the limiting reagent 36.93 H2O produced synthesis

Mg(OH)2 is the limiting reactant Mg(OH)2 + 2HCl → MgCl2 + 2H2O 1 mol Mg(OH)2 2 mol HCl 36.461 g HCl 5.87 g Mg(OH)2 x x x = 58.319 g Mg(OH)2 1 mol Mg(OH)2 1 mol HCl Given: 12.84 g HCl HCl is in excess 7.33 g HCl 1 mol Mg(OH)2 1 mol MgCl2 95.211 g MgCl2 5.87 g Mg(OH)2 x x x = 58.319 g Mg(OH)2 1 mol Mg(OH)2 1 mol MgCl2 9.58 g MgCl2 Mg(OH)2 is the limiting reactant 0.10 mol MgCl2 formed double replacement 1 mol MgCl2 9.58 g MgCl2 x 95.211 g MgCl2

Theoretical yield of H2O HI + NaOH → NaI + H2O 1 mol NaOH 1 mol H2O 18.015 g H2O 18 g NaOH x x x = 39.997 g NaOH 1 mol NaOH 1 mol H2O 8.11 g H2O produced Theoretical yield of H2O % Yield = Actual Yield Theoretical Yield x 100 7.9 % Yield = x 100 97.4% = 8.11