Question 15.16: Calculating the Volume of Solution Needed for a Reaction Inv......
Calculating the Volume of Solution Needed for a Reaction Involving Ions
A solution is 1.20 M in NO{_3} ^- ion. What volume of this solution, in milliliters, is needed to react with 10.0 g of Cu in the following redox reaction?
Cu(s) + 2 NO{_3} ^- (aq) + 4 H ^+ (aq) \longrightarrow Cu ^{2+} (aq) + 2 NO_2 (g) + 2 H_2 O(l)
Step 1 The given quantity is 10.0 g of Cu and the desired quantity is milliliters of NO{_3} ^- ion solution.
10.0 g Cu = ? mL NO{_3} ^- solution
Step 2 In terms of Figure 15.3, this is a grams-of-A to volume-of-solution B
Step 3 The dimensional analysis setup for the calculation is
10.0 \cancel{\textrm{g Cu}} \times \frac{1 \cancel{\textrm{mole Cu}}}{63.55 \cancel{\textrm{g Cu}}} \times \frac{2 \cancel{{\textrm{moles NO}_3}^- }}{1 \cancel{\textrm{mole Cu}}} \times \frac{1000 { \textrm{ mL NO}_3}^-}{1.20 \cancel{{\textrm{moles NO}_3}^-}}grams A → moles A → moles B → volume solution B
Step 4 The result, obtained by combining the numerical factors is
\frac{10.0 \times 1 \times 2 \times 1000}{63.55 \times 1 \times 1.20} mL NO{_3} ^- = 262.26069 mL NO{_3} ^- (calculator answer)
= 262 mL NO{_3} ^- (correct answer)
Answer Double Check: Is the answer reasonable? Yes. If the given amount of Cu is approximated as 0.20 of a mole the needed amount of NO{_3} ^- would be 0.40 of a mole. This NO{_3} ^- amount would require one-third (333 mL) of 1.20 M NO _3 solution. The actual molar amount of Cu is a little less than 0.20 mole. Thus, the volume of NO{_3} ^- solution needed should be less than 333 mL, which it is.
