Answer to Question #114242 in General Chemistry for Daniel

Question

How many grams of water can be produced if 3.05 L of hydrogen gas @SATP are reacted with
1.48 g of oxygen? Which is the limiting reagent?

Answer 1

V(H₂) = 3.05 L

m(O₂) = 1.48 g

2H₂ + O₂ => 2H₂O

n(H₂) = V/V_m = 3.05 L/22.4 L/mol = 0.136 mol

n(O₂) = 1.48 g/ 32 mol/L = 0.046 mol => limited reagent

n(H₂O) = n(O₂) = 0.046 mol

m(H₂O) = n(H₂O)*M(H₂O) = 0.046 mol*18 g/mol = 0.828 g