Question 4.IE: Sodium nitrite is used in the production of dyes for colorin......

Analyze

Five tasks must be performed in this problem: (1) Represent the reaction by a chemical equation in which the names of reactants and products are replaced with formulas. (2) Balance the formula equation by inspection. (3) Determine the limiting reactant. (4) Calculate the theoretical yield of sodium nitrite based on the quantity of limiting reactant. (5) Use expression (4.6) to calculate the actual yield of sodium nitrite.

percent yield = \frac{actual  yield}{theoretical  yield} × 100%             (4.6)

Solve

Obtain the unbalanced equation: Substitute formulas for names.        Na_{2}CO_{3}(aq) + NO(g) + O_{2}(g) → NaNO_{2}(aq) + CO_{2}(g)

Obtain the balanced equation: Begin by noting that there must be 2 NaNO_{2} for every Na_{2}CO_{3}.            Na_{2}CO_{3}(aq) + NO(g) + O_{2}(g) → 2  NaNO_{2}(aq) + CO_{2}(g)

To balance the N atoms requires 2 NO on the left side:        Na_{2}CO_{3}(aq) + 2  NO(g) + O_{2}(g) → 2  NaNO_{2}(aq) + CO_{2}(g)

To balance the O atoms requires \frac{1}{2} O_{2} on the left, resulting in six O atoms on each side:          Na_{2}CO_{3}(aq) + 2  NO(g) + \frac{1}{2} O_{2}(g) → 2  NaNO_{2}(aq) + CO_{2}(g)

Multiply coefficients by 2 to make all of them integers in the final balanced equation.         2 Na_{2}CO_{3}(aq) + 4  NO(g) + O_{2}(g) → 4  NaNO_{2}(aq) + 2  CO_{2}(g)

Determine the limiting reactant: The problem states that oxygen is in excess. To determine the limiting reactant, the number of moles of Na_{2}CO_{3}, requiring molarity as a conversion factor, must be compared with the number of moles of NO, requiring a conversion factor based on the molar mass of NO.          ? mol Na_{2}CO_{3} = 0.225  L  soln × \frac{1.50  mol  Na_{2}CO_{3}}{1  L  soln}

= 0.338 mol Na_{2}CO_{3}

? mol NO = 22.1 g NO × \frac{1  mol  NO}{30.01  g  NO} = 0.736 mol NO

From the balanced equation, the stoichiometric mole ratio is            4 mol NO : 2 mol Na_{2}CO_{3} = 2 : 1

The available mole ratio is          0.736 mol NO : 0.338 mol Na_{2}CO_{3} = 2.18 : 1

Because the available mole ratio exceeds 2 : 1, NO is in excess and Na_{2}CO_{3} is the limiting reactant.

Determine the theoretical yield of the reaction: This calculation is based on the amount of Na_{2}CO_{3}, the limiting reactant.          ? g NaNO_{2} = 0.338  mol  Na_{2}CO_{3} × \frac{4  mol  NaNO_{2}}{2   mol  Na_{2}CO_{3}} × \frac{69.00  g  NaNO_{2}}{1  mol  NaNO_{2}} = 46.6  g  NaNO_{2}

Determine the actual yield: Use expression (4.6), which relates percent yield (95.0%), theoretical yield (46.6 g NaNO_{2}), and the actual yield.          actual yield = \frac{percent  yield  ×  theoretical  yield}{100\%}

= \frac{95.0\%  ×  46.6  g  NaNO_{2}}{100\%} = 44.3  g  NaNO_{2}

Assess

After solving a multistep problem, it is important to check over your work. In this example, we should first double-check that the chemical equation is properly balanced. There are 4 Na’s on each side, 2 C’s, 12 O’s, and 4 N’s. We can double-check that we correctly identified the limiting reactant by using a different approach. Because 0.338 mol Na_{2}CO_{3} would yield 0.676 mol NaNO_{2} (in the presence of excess NO and O_{2}) and 0.736 mol NO would yield 0.736 mol NaNO_{2} (in the presence of excess Na_{2}CO_{3}  and  O_{2}), we conclude that Na_{2}CO_{3} must be the limiting reactant; the limiting reactant is the one that limits the amount of product obtained. In checking the calculation of the mass of NaNO_{2} obtained, we notice that the units work out properly. To check the final step, we can use our final answer to calculate the percent yield for the experiment: (44.3/46.6) × 100% = 95%. This is the correct result for the percent yield.