Exercises
Exercise 1. What mass of NaCl is needed to precipitate Ag+ ions from 25.00 mL of 0.250 M AgNO3 solution?
First, write a balanced chemical equation.
NaCl (aq) + AgNO3(aq) → AgCl (s) + NaNO3 (aq)
Calculate the number of moles of Ag+ ion. From the equation stoichiometry there is a 1:1 mole ratio of Ag+:NaCl.
\(\displaystyle moles\;AgNO_3\;=\;0.02500\;L\;\times\frac{0.250\;mol}{L}\;=\;\mathbf{0.00625\;mol\;AgNO_3}\)There is 0.00625 mol of Ag+ ion in 0.00625 mol AgNO3.
\(\displaystyle 0.00625\;mol\;AgNO_3\;\times\frac{1\;mol\;NaCl}{1\;mol\;AgNO_3}\;=\;\mathbf{0.00625\;,mol\;NaCl}\)The molar mass of NaCl is 58.44 g/mol.
\(\displaystyle 0.00625\;mol\;NaCl\times\frac{58.44\;g\;NaCl}{1\;mol\;NaCl}\;=\;\mathbf{0.365\;g\;NaCl}\)
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Exercise 2. How many mL of 0.125 M HCl are required to react with 45.00 mL of 0.150 M Ba(OH)2?
First, write a balanced equation for the reaction.
2 HCl (aq) + Ba(OH)2 (aq) → BaCl2 (aq) + 2 H2O (l)
Calculate the number of moles of Ba(OH)2. Then use the equation stoichiometry to calculate the number of moles of HCl.
\(\displaystyle 0.04500\;L\;\times\;0.150\;M\;=\;\mathbf{0.00675\;mol\;Ba(OH)_2}\)
\(\displaystyle 0.00675\;mol\;Ba(OH)_2\times\frac{2\;mol\;HCl}{1\;mol\;Ba(OH)_2}\;=\;\mathbf{0.0135\;mol\;HCl}\)
Finally, determine the volume in L and then convert to mL.
\(\displaystyle \frac{0.0135\;mol\;HCl}{0.125\;\;HCl}\;=\;\mathbf{0.108\;L\;HCl\;=\;108\;mL\;HCl}\)
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Exercise 3. Acetic acid, CH3COOH, is the main acid component of vinegar. If 3.56 mL of vinegar requires 38.25 mL of 0.125 M NaOH for reaction, how many grams of acetic acid are in 545 mL of vinegar?
The equation is balanced. Find the number of moles of NaOH and then use the equation stoichiometry to calculate the moles of acetic acid in 3.56 mL of vinegar.
\(\displaystyle 0.03825\;L\;\times\frac{0.125\;mol}{L}\;=\;0.00478\;mol\;NaOH\)
\(\displaystyle 0.00478\;mol\;NaOH\times\frac{1\;mol\;CH_3COOH}{1\;mol\;NaOH}\;=\;0.00478\;mol\;CH_3COOH\)
The molar mass of CH3COOH is 60.052 g/mol. Convert moles to grams.
\(\displaystyle 0.00478\;mol\;CH_3COOH\times\frac{60.052\;g}{1\;mol}\;=\;\mathbf{0.287\;g\;CH_3COOH}\)There are 0.287 g of CH33COOH in 3.56 mL of vinegar. Calculate the number of grams of CH3COOH in 545 mL of vinegar.
\(\displaystyle 545\;mL\times\frac{0.287\;g\;CH_3COOH}{3.56\;mL}\;=\;\mathbf{43.9\;g\;CH_3COOH}\)
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Exercise 4. Consider the following chemical equation:
How many grams of BaCl2 are formed if 125.00 mL of 0.355 M AlCl3 is reacted with 150.00 mL of 0.296 M Ba(OH)2?
First, balance the chemical equation:
2 AlCl3 (aq) + 3 Ba(OH)2 (aq) → 2 Al(OH)3 (s) + 3 BaCl2 (aq)
Determine the number of moles of AlCl3 and Ba(OH)2
\(\displaystyle mol\;AlCl_3\;=\;0.12500\;L\times\;0.355\;mol\;=\;0.4438\;mol\;AlCl_3\)
\(\displaystyle mol\;Ba(OH)_2\;=\;0.15000\;L\times\;0.296\;mol\;=\;0.0444\;mol\;Ba(OH)_2\)
Find the limiting reactant.
\(\displaystyle 0.4438\;mol\;AlCl_3\times\frac{3\;mol\;BaCl_2}{2\;mol\;AlCl_3}\;=\;
0.6657\;mol\;BaCl_2\)
\(\displaystyle 0.0444\;mol\;Ba(OH)_2\times\frac{3\;mol\;BaCl_2}{3\;mol\;Ba(OH)_2}\;=\;
0.0444\;mol\;BaCl_2\)
The limiting reactant is Ba(OH)2. The molar mass of BaCl2 is 208.23 g/mol.
\(\displaystyle 0.0444\;mol\;BaCl_2\times\frac{208.23\;g\;BaCl_2}{1\;mol\;BaCl_2}\;=\;
\mathbf{9.25\;g\;BaCl_2}\)