Question

Nitric acid reacts with silver metal 4hno3 + 3ag = no + 2h2o + 3agno3. calculate the number of grams of no formed when 10.8 g of ag reacts with 12.6g of hno3

Answer 1

The balanced equation for the reaction is;
2HNO₃ + Ag ---> NO + 2H₂O + AgNO₃
Stoichiometry of HNO₃ to Ag is 2:1
we need to first find the limiting reactant
number of HNO₃ moles - 12.6 g / 63 g/mol  = 0.2 mol
number of Ag moles - 10.8 g / 108 g/mol = 0.1 mol
0.2 moles of HNO₃ react with 0.1 mol of Ag, this is in the 2:1 molar ratio which means that both reactants react fully in the reaction.
then molar ratio of Ag to NO is 1:1
number of Ag moles equivalent to number of NO moles 
Number of NO moles - 0.1 mol
Mass of NO - 0.1 mol x 30 g/mol = 3 g
mass of NO produced is 3 g