If the reaction is allowed to proceed with 8 moles of sodium nitrite and 6 moles of sulfuric acid, which reagent is the limiting reagent?
The mole ratio on the reactant side of equation (1) shows that for every 1 mole of sulfuric acid, 2 moles of sodium nitrite are needed. Therefore, for all 6 moles of sulfuric acid to be used, 12 moles of sodium nitrite would be needed. Since there are only 8 moles of sodium nitrite, it must be the limiting reagent.
Example:
In equation (1), how much excess reagent is there?
The 8 moles of sodium nitrite can react with 4 moles of sulfuric acid, leaving 2 moles of sulfuric acid as the excess reagent.