Honors Chemistry I
Stoichiometry Practice Test
Note: This practice test has more questions on it than you will see on the actual test. I have tried to include
a large variety of questions so that you will be prepared for Thursday's test!
I.
A student dissolves 3.01 grams of solid copper metal in an excess of 6.00 M nitric acid, producing
gaseous nitric oxide (NO), cupric nitrate, and water. Once the gas was removed, the blue solution was
heated to drive off all of the water and 8.70 grams of a blue precipitate were recovered. The reaction
was conducted at STP conditions. Answer the following questions based on this information.
A. Write the balanced equation for this reaction. [Hint: the sum total of all coefficients adds up to 20.]
3Cu + 8HNO3  2NO + 3Cu(NO3)2 + 4H2O (REDOX!!!)
B. What volume of nitric oxide should have been formed?
3.01 g Cu x (1 mol/63.55 g) x (2 NO/3 Cu) x (22.41 L/1 mol) = .707 L
C. What is the mass of water produced?
3.01 x (1 mol/63.55 g) x (4H2O/3Cu) x 18.016 g/1 mol) = 1.14 g
D. How many moles of nitric acid were needed initially?
3.01 g Cu x (1 mol/63.55 g) x ( 8HNO3/3Cu) = 0.126 mol
E. How many grams of cupric nitrate should have been produced, assuming 100% yield?
3.01 g (1 mol/63.55 g) x ( 3Cu(NO3)2/ 3Cu) x (187.57 g/1 mol) = 8.88 g
F. What is the percent yield for this reaction?
8.70/8.88 x 100 = 98.0 %
G. If the student wanted to add exactly the stoichiometric amount of acid needed to dissolve the copper
metal, given that the concentration of the acid was 6.00 M, what volume of acid would have been
needed?
6.00 M = .126 / x
x = 0.021
-2II.
When 120.0 ml of a 3.00 M solution of nickel (II) chlorate are mixed with 150.0 ml of a 2.00 M
solution of sodium phosphate, 45.0 grams of a bright green precipitate were recovered. Answer the
following questions based on this information.
A. Write the balanced equation for this reaction.
3Ni(ClO3)2 + 2Na3PO4  6NaClO3 + Ni3(PO4)2
B. Write the net ionic equation for the reaction.
3Ni+2 + 2PO4-3  Ni3(PO4)2
C. How many moles of each reactant were there?
Ni(ClO3)2 = 3.00 x .1200 = .360 mol
Na3PO4 = 2.00 x .1500 = .300 mol
D. What is the limiting reagent?
.360 mol Ni(ClO3)2 x (2 Na3PO4/ 3 Ni(ClO3)2 ) = 0.240 mol needed
.300 mol Na3PO4 x ( 3 Ni(ClO3)2/2 Na3PO4 ) = 0.45 mol needed
E. How many moles of excess reagent were left unreacted?
.300 mol - .240 mol = .060 mol left unreacted
F. What is the precipitate?
nickel(II) phosphate
G. How many grams of precipitate should have been formed theoretically?
.360 mol Ni(ClO3)2 x (1 Ni3(PO4)2/ 3Ni(ClO3)2) x (366.04 g/1 mol) = 43.9 g
H. What is the percent yield for this reaction? [Be careful!]
45.0/43.9 x 100 = 103%
I. What is the [Ni+2] in the final solution?
0M
J. What is the [PO4-3] in the final solution?
0.060 mol /.270 L = .22 M
K. What is the [ClO3-1] in the final solution?
.360 mol x (2/1) = .720
.720 mol/.270 L = 2.67 M
L. What is the [Na+1] in the final solution?
.300 x (3/1) = .900 mol
.900 mol/.270 L = 3.33 M
M. How many more moles of the limiting reagent would need to be added to produce stoichiometric
amounts?
.450-.360 = .090 mol