Question

Given the same reaction Na2S(aq) + 2 AgNO3(aq) → Ag2S(s) + 2 NaNO3(aq)

​a. If I react 23.7 grams of Na2S with 108 grams of AgNO3, how many grams of NaNO3 do I make?

​b. How much excess reactant (grams) is left over?

​

Answer 1

The balanced equation is Na S (aq) + 2 AgNO3 (aq) → Ag S (s) + 2 NaNO3 (aq) Number of moles of Na S = mass/mol.wt. = 23.7 g/(78.04 g/mol) = 0.30369 mol Number of moles of AgNO3 = mass/mol.wt. = 108 g/(169.87 g/mol) = 0.63578 mol According to balanced equation, 1 mole of Na S reacts with 2 moles of AgNO3 So, 0.30369 mol of Na,s would react with 0.30369 mol * 2/1 = 0.607381 mol of AgNO3 But, the number of moles of AgNO3 present more than required. That is, AgNO3 is the excess reagent. Na,S is the Limiting reagent. According to balanced equation, 1 mole of Na2S produces 2 mole of NaNO3 So, 0.30369 mol of Na,S would produce 0.30369 mol * 2/1 = 0.607381 mol of NaNO2 Mass of NaNO3 = 0.607381 mol * 84.99 g.mol = 51.6213 g = 51.62 g Number of moles of AgNO, present in excess = 0.63578 -0.607381 = 0.028399 mol Mass of AgNO3 (excess reagent) = 0.028399 mol * 169.87 g/mol = 4.82 g