Inactive Tutor answered • 05/20/21
BCE: 2 NaCl + Pb(NO3)2 ---> PbCl2 + 2 NaNO3
How many grams of sodium nitrate can be produced with 60 grams of sodium chloride and 55 grams of lead IV nitrate?
60g NaCl x (mol / 58.44g) = 1.03 mol NaCl x (2 mol NaNO3 / 2 mol NaCl) = 1.03 mol NaNO3
55g Pb(NO3)2 x (mol / 331.2g) = 0.17 mol Pb(NO3)2 x (2 mol NaNO3 / 1 mol Pb(NO3)2) = 0.34 mol NaNO3
NaCl -- 1.03 mol NaNO3 x (84.9947g / mol) = 87.5g NaNO3
Pb(NO3)2 -- 0.34 mol NaNO3 x (84.9947g / mol) = 28.9g NaNO3
Answer: 29g sodium nitrate
Which reactant is limiting?
Answer: lead IV nitrate (55g)
How many grams of the excess reactant will be left after the reaction?
0.17 mol Pb(NO3)2 x (2 mol NaCl / 1 mol Pb(NO3)2) = 0.34 mol NaCl
0.34 mol NaCl x (58.44g / mol) = 19.9g NaCl
60g NaCl - 19.9g NaCl = 40.1g NaCl
Answer: 40g NaCl
**This last part is what can be confusing for some students. You have to take the amount of excess reactant that was put into the reaction and subtract from that the amount that was actually used. You calculate the amount that was actually used by converting the the moles of limiting reactant to moles of excess reactant and then converting moles of excess reactant to grams.
