Question

Which is the limiting reactant when 5.00 g of H₂ and 10.0 g of O₂ react and form water?

How many grams of water are formed and what mass of excess reagent remains?

Answer 1

the balanced equation is as follows

2 H2 + O2 -------------> 2 H2O

no of moles = mass / moles

for H2 = 5.00 g / 2 g/mol => 2.5 moles

for O2 = 10.0g / 32g/mol => 0.3125moles

2 mol H2 --------------------> 2 mol H2O

2.5 mol ------------------------>?

=> 2.5 * 2 / 2 => 2.5 moles of H2O

1 mol O2 ----------------> 2 mol H2O

0.3125 mol ------------------->?

=> 0.3125 * 2 / 1 => 0.625 moles of H2O

limiting reactant is O2

mass of H2O formed => 0.625 moles * 18 g/mol => 11.25 grams

excess reactant is H2

excess moles => 2.5 - 0.625 => 1.875 moles

excess mass => 1.875 * 2 => 3.75 grams