Download Stoichiometric calculations

Chemistry 101
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Stoichiometric calculations
All problems are based on the reaction of magnesium chloride with sodium
hydroxide.
1. Predict the products.
Mg(OH)2 + NaCl
2. Write a balanced chemical equation for the reaction.
MgCl2 + 2 NaOH
Mg(OH)2 + 2 NaCl
3. Calculate the percent composition of all the elements in the magnesiumcontaining product.
Mg(OH)2
molar mass = 24.3 + 2 (16.0) + 2 (1.0) = 58.3 g/mole
% composition =
% Mg 
g of element
X 100
Total g of compound
24.3g
 100  41.7%
58.3g
% O = 2(16.0g) X 100 = 54.9%
58.3g
%H = 2(1.0g) X 100 = 3.4%
58.3g
4. How many moles of each product could be formed from 3.72 moles of sodium
hydroxide?
3.72molesNaOH 
2molesNaCl
 3.72molesNaCl
2molesNAOH
3.72molesNaOH 
1moleMg (OH ) 2
 1.86molesMg (OH ) 2
2molesNaOH
5. How many grams of each product could be formed from 7.32 grams of
magnesium chloride?
Steps
1. gA
2. moles A
3. moles B
moles A
moles B
gB
Product #1
Molar mass MgCl2 = 24.3 + 2(35.5) = 95.3g MgCl2
1mole MgCl2
1moleMgCl 2
7.32 gMgCl 2 
 0.0768molesMgCl 2
95.3gMgCl 2
0.0768molesMgCl 2 
1moleMg (OH ) 2
 0.0768molesMg (OH ) 2
1moleMgCl 2
Molar mass of Mg(OH)2 = 24.3 + 2(16.0) + 2(1.0) = 58.3g Mg(OH)2
1mole Mg(OH)2
58.3gMg (OH ) 2
0.0768molesMg (OH ) 2 
 4.48gMg (OH ) 2
1moleMg (OH ) 2
Product #2
Molar mass
7.32 gMgCl 2 
1moleMgCl 2
 0.0768molesMgCl 2
95.3gMgCl 2
0.0768molesMgCl 2 
2molesNaCl
 0.153molesNaCl
1moleMgCl 2
Molar mass NaCl = 23.0 + 35.5 = 58.5 g NaCl
1mole NaCl
58.3gNaCl
0.153molesNaCl 
 8.96 gNaCl
1moleNaCl