Question

2. What is the limiting reactant when 126 g of Ag reacts with 142 g of HNO3? What is the heoretical yield in grams of AgNO3? Ag(s) + HNO3 (aq) → AgNO3(aq) + NO(g) + H2O)

Answer 1

From the balanced chemical equation, it is clear that 1 mole of Ag requires 1 mole of HNO3. Let us calculate the number of moles of Ag and HNO3 used in the reaction. The molar masses of Ag and HNO3 are, 107.87 g/mol and 63 g/mol, repsectively.

The number of moles of Ag = Weight of Ag in g / Molar mass of Ag in g/mol

= 126 g / 107.87 g/mol = 1.17 moles

Number of moles of HNO3 = Weight of HNO3 in g / Molar mass of HNO3 in g/mol

= 142 g / 63 g/mol = 2.25 mol

From the calculated values of number of moles for Ag and HNO3, it is clear that HNO3 has been used in excess. Therefore, the limiting reagent is Ag.

Therefore the yield of AgNO3 formed would be dependent on the number of moles of Ag used in the reaction.

From the balanced equation, it can be seen that 1 moles of Ag gives 1 moles of AgNO3.

As 1.17 moles of of Ag is used in the reaction, it should give 1.17 moles of AgNO3. The molar mass of AgNO3 is 169.87 g/mol. The mass of AgNO3 formed can be calculated using following formula,

Mass of AgNO3 = number of moles of AgNO3 x molar mass of AgNO3

= 1.17 mol x 169.87 g/mol

= 198.74 g

Therefore, the theoretical yield og AgNO3 in gram will be 198.74 g