Equation Stoichiometry
Worksheet-1
1. When potassium chlorate is heated, it decomposes to potassium chloride and
oxygen gas.
a) Write a balanced molecular equation showing this reaction. Include
phases.
2KClO3(s)  2KCl(s) + 3O2(g)
b) If 0.500 grams of potassium chlorate is decomposed, calculate the moles
of potassium chloride and moles of oxygen gas that will be produced.
0.500 g KClO3 x (1 mol KClO3/122.6 g KClO3) x (2 mol KCl/2 mol KClO3) =
0.00408 mol KCl
0.500 g KClO3 x (1 mol KClO3/122.6 g KClO3) x (3 mol O2/2 mol KClO3) =
0.00612 mol O2
c) Calculate the masses of potassium chloride and oxygen gas that will be
produced.
0.00408 mol KCl x (74.6 g KCl/1 mol KCl) = 0.304 g KCl
0.00612 mol O2 x (32.0 g O2/1 mol O2) = 0.196 g O2
Or 0.500 = 0.304 + X, x = 0.196 g O2
2. A 4.25 mole sample of molten aluminum chloride undergoes electrolysis.
a) write a balanced molecular equation showing this reaction
2AlCl3(l)  2Al(s) + 3Cl2(g)
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b) calculate the masses of products formed.
4.25 mol AlCl3 x (2 mol Al/2 mol AlCl3) x (27.0 g Al/1 mol Al) = 115 g Al
4.25 mol AlCl3 x (3 mol Cl2/2 mol AlCl3) x (71.0 g Cl2/1 mol Cl2) = 453 g Cl2
3. A sample of magnesium metal reacts pure nitrogen gas.
a) write a balanced molecular equation for this reaction
3Mg(s) + N2(g)  Mg3N2(s)
b) calculate the masses of each of the reactants that would be needed to
produce exactly 0.400 mole of magnesium nitride.
0.400 mol Mg3N2 x (3 mol Mg/1 mol Mg3N2) x (24.3 g Mg/1 mol Mg) = 29.2 g Mg
0.400 mol Mg3N2 x (1 mol N2/1 mol Mg3N2) X (28.0 g N2/1 mol N2) = 11.2 g N2
4. A sample of zinc metal is placed in a solution of hydrochloric acid.
a) write a balanced molecular equation for this reaction
Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g)
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Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g)
If 2.80 g of zinc reacts with excess acid,
b) calculate the moles of hydrogen gas produced
2.80 g Zn x (1 mol Zn/65.4 g Zn) x (1 mol H2/1 mol Zn) = 0.0428 mol H2
c) calculate the mass of hydrogen gas produced
0.0428 mol H2 x (2.0 g H2/1 mol H2) = 0.0856 g H2
d) if the solution is allowed to evaporate, determine the mass of salt that will
be found in the bottom of the beaker.
2.80 g Zn x (1 mol Zn/65.4 g Zn) x (1 mol ZnCl2/1 mol Zn) X (136.4 g
ZnCl2/1 mol Zn Cl2) = 5.84 g ZnCl2
5. Exactly 4.75 grams of copper(II)chloride in solution reacts with excess sodium
hydroxide solution.
a) Write a balanced molecular equation for this reaction
CuCl2(aq) + 2NaOH(aq)  Cu(OH)2(s) + 2NaCl(aq)
b) Calculate the moles of precipitate that will form
4.75 g CuCl2 x (1 mol CuCl2/134.5 g CuCl2) X (1 mol Cu(OH)2/1 mol CuCl2) = 0.0353
mol Cu(OH)2
c) Calculate the mass of precipitate that will form
0.0353 mol Cu(OH)2 x (97.5 g Cu(OH)2/1 mol Cu(OH)2 ) = 3.44 g Cu(OH)2
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