Question 3.7: FINDING THE MASS OF ONE REACTANT, GIVEN THE MASS OF ANOTHER ...
FINDING THE MASS OF ONE REACTANT, GIVEN THE MASS OF ANOTHER
Aqueous solutions of sodium hypochlorite (NaOCl), best known as household bleach, are prepared by reaction of sodium hydroxide with chlorine. How many grams of NaOH are needed to react with 25.0 g of Cl_{2}?
2 NaOH(aq) + Cl_{2}(g) → NaOCl(aq) + NaCl(aq) + H_{2}O(l)STRATEGY
Finding the relationship between numbers of reactant formula units always requires working in moles, using the general strategy outlined in Figure 3.1.
First, find out how many moles of Cl_{2} are in 25.0 g of Cl_{2}. This gram-to-mole conversion is done in the usual way, using the molar mass of Cl_{2} (70.9 g/mol) as the conversion factor:
25.0 \cancel{g Cl_{2}} × \frac{1 mol Cl_{2}}{70.9 \cancel{g Cl_{2}}} = 0.353 mol Cl_{2}Next, look at the coefficients in the balanced equation. Each mole of Cl_{2} reacts with 2 mol of NaOH, so 0.353 mol of Cl_{2} reacts with 2 × 0.353 = 0.706 mol of NaOH. With the number of moles of NaOH known, carry out a mole-to-gram conversion using the molar mass of NaOH (40.0 g/mol) as a conversion factor to find that 28.2 g of NaOH is required for the reaction:
Grams of NaOH = 0.353 \cancel{mol Cl_{2}} × \frac{2 \cancel{mol NaOH}}{1 \cancel{mol Cl_{2}}} × \frac{40.0 gNaOH}{1 \cancel{mol NaOH}}
= 28.2 g NaOH
The problem can also be worked by combining the steps and setting up one large equation:
Grams of NaOH = 25.0 \cancel{g Cl_{2}} × \frac{1 \cancel{mol Cl_{2}}}{70.9 \cancel{g Cl_{2}}} × \frac{2 \cancel{mol NaOH}}{1 \cancel{mol Cl_{2}}} × \frac{40.0 g NaOH}{1 \cancel{mol NaOH}}
= 28.2 gNaOH
BALLPARK CHECK
The molar mass of NaOH is about half that of Cl_{2}, and 2 mol of NaOH is needed per 1 mol of Cl_{2}. Thus, the needed mass of NaOH will be similar to that of Cl_{2}, or about 25 g.
Household bleach is an aqueous solution of NaOCl, made by reaction of NaOH with Cl_{2}