QuestionJune 20, 2025

Solutions of sodium hypochlorite. NaClO are sold as a bleach (such as Clorox). They are prepared by the reaction of chlorine with sodium hydroxide. 2NaOH(aq)+Cl_(2)(g)arrow NaCl(aq)+NaClO(aq)+H_(2)O(l) If you have 1.29 mol of NaOH in solution and 1.60 mol of Cl_(2) gas available to react,which is the limiting reactant? NaOH Cl_(2) How many moles of NaClO(aq) could be obtained? square mol

Solutions of sodium hypochlorite. NaClO are sold as a bleach (such as Clorox). They are prepared by the reaction of chlorine with sodium hydroxide. 2NaOH(aq)+Cl_(2)(g)arrow NaCl(aq)+NaClO(aq)+H_(2)O(l) If you have 1.29 mol of NaOH in solution and 1.60 mol of Cl_(2) gas available to react,which is the limiting reactant? NaOH Cl_(2) How many moles of NaClO(aq) could be obtained? square mol
Solutions of sodium hypochlorite. NaClO are sold as a bleach (such as Clorox). They are prepared by the reaction of chlorine with sodium hydroxide. 2NaOH(aq)+Cl_(2)(g)arrow NaCl(aq)+NaClO(aq)+H_(2)O(l) If you have 1.29 mol of NaOH in solution and 1.60 mol of Cl_(2) gas available to react,which is the limiting reactant? NaOH Cl_(2) How many moles of NaClO(aq) could be obtained? square mol

Solution
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Answer

0.645 mol Explanation 1. Determine the stoichiometric ratio The balanced equation is 2 \text{NaOH} + \text{Cl}_2 \rightarrow \text{NaCl} + \text{NaClO} + \text{H}_2O. The stoichiometric ratio of NaOH to Cl_2 is 2:1. 2. Calculate moles required for reaction For 1.60 mol of Cl_2, you need 2 \times 1.60 = 3.20 mol of NaOH. 3. Identify the limiting reactant You have only 1.29 mol of NaOH, which is less than 3.20 mol needed. Thus, NaOH is the limiting reactant. 4. Calculate moles of NaClO produced From the balanced equation, 2 mol of NaOH produces 1 mol of NaClO. Therefore, 1.29 mol of NaOH will produce \frac{1.29}{2} = 0.645 mol of NaClO.

Explanation

1. Determine the stoichiometric ratio<br /> The balanced equation is $2 \text{NaOH} + \text{Cl}_2 \rightarrow \text{NaCl} + \text{NaClO} + \text{H}_2O$. The stoichiometric ratio of NaOH to Cl$_2$ is 2:1.<br /><br />2. Calculate moles required for reaction<br /> For 1.60 mol of Cl$_2$, you need $2 \times 1.60 = 3.20$ mol of NaOH.<br /><br />3. Identify the limiting reactant<br /> You have only 1.29 mol of NaOH, which is less than 3.20 mol needed. Thus, NaOH is the limiting reactant.<br /><br />4. Calculate moles of NaClO produced<br /> From the balanced equation, 2 mol of NaOH produces 1 mol of NaClO. Therefore, 1.29 mol of NaOH will produce $\frac{1.29}{2} = 0.645$ mol of NaClO.
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