3.67 g of Na react with 8.00 g of H2o write the formula for the reactant that is limiting reactant. how many grams of h2 are formed? how much of the excess reactant in grams remains after the reaction

Answer 1

2Na+2H2O--> 2NaOH+H2

3.67gNa/1 × 1molNa/23gNa × 2molH2O/2molNa × 18.02gH2O/1molH2O=2.88 g H2O needed

limiting reactant=Na

excess H2O=8.00g-2.88g=5.12g H2O in excess

grams of H2 formed from limiting reactant:

3.67gNa/1 × 1molNa/23gNa × 2molH2/2molNa × 2.02gH2/1molH2=0.322 g H2