excess sodium hydroxide is added to a solution containing 4.6 grams of magnesium chloride. a reaction takes…

excess sodium hydroxide is added to a solution containing 4.6 grams of magnesium chloride. a reaction takes place according to this equation: 2naoh(aq) + mgcl₂(aq) → 2nacl(aq) + mg(oh)₂(s). the magnesium hydroxide produced by the reaction was collected and weighed. if the mass of the magnesium hydroxide was 2.7 grams, what was the percent yield? use the periodic table. a. 48% b. 59% c. 61% d. 96%

excess sodium hydroxide is added to a solution containing 4.6 grams of magnesium chloride. a reaction takes place according to this equation: 2naoh(aq) + mgcl₂(aq) → 2nacl(aq) + mg(oh)₂(s). the magnesium hydroxide produced by the reaction was collected and weighed. if the mass of the magnesium hydroxide was 2.7 grams, what was the percent yield? use the periodic table. a. 48% b. 59% c. 61% d. 96%

Answer

Explanation:

Step1: Calculate molar mass of MgCl₂

Molar mass of MgCl₂: $M_{MgCl_2}=24.3 + 2\times35.5=95.3$ g/mol

Step2: Calculate moles of MgCl₂

Moles of MgCl₂, $n_{MgCl_2}=\frac{m_{MgCl_2}}{M_{MgCl_2}}=\frac{4.6}{95.3}\approx0.0483$ mol

Step3: Determine moles of Mg(OH)₂ from stoichiometry

From the reaction $2NaOH(aq)+MgCl_2(aq)\to2NaCl(aq)+Mg(OH)2(s)$, the mole - ratio of MgCl₂ to Mg(OH)₂ is 1:1. So moles of Mg(OH)₂ produced theoretically, $n{Mg(OH)2}^{theo}=n{MgCl_2}=0.0483$ mol

Step4: Calculate molar mass of Mg(OH)₂

Molar mass of Mg(OH)₂: $M_{Mg(OH)_2}=24.3+(2\times(16 + 1))=58.3$ g/mol

Step5: Calculate theoretical mass of Mg(OH)₂

Theoretical mass of Mg(OH)₂, $m_{Mg(OH)2}^{theo}=n{Mg(OH)2}^{theo}\times M{Mg(OH)_2}=0.0483\times58.3\approx2.82$ g

Step6: Calculate percent yield

Percent yield, $\text{Percent Yield}=\frac{m_{Mg(OH)2}^{actual}}{m{Mg(OH)_2}^{theo}}\times100=\frac{2.7}{2.82}\times100\approx96%$

Answer:

D. 96%