How many liters of NO(g) are produced when 0.85L of 2.00 M HNO3 solution is added to 216 of Ag according to the equation.

Question

Which reactant is the limiting reactant? 3 Ag(s) + 4 HNO3(aq) 3 AgNO3(aq) + NO(g) + 2 H2O(l)

Answer 1

3 Ag(s) + 4 HNO3(aq) 3 AgNO3(aq) + NO(g) + 2 H2O(l)

Use dimensional analysis and stoichiometry to solve:

moles HNO₃ present = 0.85 L x 2.00 mol / L = 1.70 moles HNO₃

moles Ag present = 216 g Ag x 1 mol Ag / 108 g = 2.00 moles Ag

An easy way to find the limiting reactant is to divide the moles of each reactant by its coefficient and the lower value represents the limiting reactant.

In this case, we have...

1.70 mol HNO₃ / 4 = 0.425

2.00 mol Ag / 3 = 0.66

HNO₃ is limiting

Liters of NO gas formed = 1.70 mol HNO₃ x 1 mol NO / 4 mol HNO₃ x 22.4 L / mol = 9.52 L NO (assuming STP)