Chemistry I-Honors
Stoichiometry - Limiting Reagents
Worksheet #1 Solution Set
I. Joe Student mixes 20.0 ml of a 3.00 M solution of silver nitrate with 15.0 ml of a 2.00 M solution of sodium phosphate. He collects and dries the precipitate formed, then weighs the precipitate and finds that it has a mass of 8.00 grams. Answer the following questions based on this information.
A. Write the balanced equation for this reaction.
3 AgNO3(aq) + Na3PO4(aq) à Ag3PO4(ppt.) + 3 NaNO3(aq)
B. How many moles of each reactant were there?
(0.0200 L)(3.00 M) = 0.0600 mol AgNO3
(0.0150 L)(2.00 M) = 0.0300 mol Na3PO4
C. What is the limiting reagent?
0.0300 mol Na3PO4 (3 mol AgNO3 / 1 mol Na3PO4) = 0.0900 mol AgNO3 needed
only have 0.0600 mol AgNO3 so it is the limiting reagent
D. How many moles of excess reagent were left unreacted?
0.0600 mol L.R. (1 mol E.R. / 3 mol L.R.) = 0.0200 mol E.R. needed
0.0300 mol - 0.0200 mol = 0.0100 mol xcs
E. What is the precipitate? Ag3PO4(ppt.)
F. How many moles of precipitate should have been formed theoretically?
0.0600 mol L.R. (1 mol ppt. / 3 mol L.R.) = 0.0200 mol ppt.
[ = 8.37 grams ppt.]
G. What is the percent yield for this reaction?
8.00
------x 100 = 95.6%
8.37
H. What is the concentration of each of the spectator ions (two of them)?
(0.0300 mol) x 3 0.0600 mol
[Na+1] = ------= 2.57 M [NO3-1] = ------= 1.71 M
0.0350 L 0.0350 L
I. What is the concentration of the excess reagent ion?
[PO4-3] = 0.0100 mol / 0.0350 L = 0.286 M
J. How many more moles of the limiting reagent would need to be added to produce stoichiometric amounts?
from C - would need 0.0300 more moles of the L.R.
II. Josephine Student (Joe's twin sister) runs another experiment. She reacts 4.00 grams of solid copper metal with 10.0 ml of concentrated nitric acid. Concentrated nitric acid is 15.8 M. The three products are cupric nitrate, nitric oxide (NO), and water. She collects the NO gas and finds it has a volume of 805 ml at STP conditions. Answer the following questions based on this information.
A. Write the balanced equation for this reaction.
3 Cu(s) + 8 HNO3(aq) à 3 Cu(NO3)2(aq) + 2 NO(g) + 4 H2O(liq)
B. How many moles of each reactant were there?
4.00 g. Cu / 63.5 g/mol = 0.0630 mol Cu (0.0100 L)(15.8 M) = 0.158 mol acid
C. What is the limiting reagent?
0.158 moles acid (3 mol Cu / 8 mol acid) = 0.0593 moles needed Cu
there are 0.0630 moles of Cu - or an excess - so the acid is the L.R.
D. How many moles of excess reagent were left unreacted?
0.0630 moles - 0.0593 moles = 0.0037 moles xcs
E. How many grams of water could be produced, assuming 100% yield?
0.158 mol LR ( 4 mol water / 8 mol LR)(18.0 g water / 1 mol water) = 1.42 g H2O
F. How many liters of nitric oxide gas should have been formed theoretically?
0.158 mol LR ( 2 mol NO / 8 mol LR)(22.4 L NO / 1 mol NO) = 0.885 L NO
G. What is the percent yield for this reaction?
(805 mL / 885 mL) x 100 = 90.96 = 91.0%
H. How many more moles of the limiting reagent would need to be added to produce stoichiometric amounts?
0.0630 moles Cu (8 moles acid / 3 moles Cu) = 0.168 moles acid needed to react
0.168 mol - 0.158 mol = 0.010 mole needed