Question

Select the correct answer.

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 1

The answer is D ; 96%

Step-by-step explanation:

Answer 2

Answer : The correct option is, 96%

Solution : Given,

Mass of magnesium chloride = 4.6 g

Molar mass of magnesium chloride = 95.2 g/mole

Molar mass of magnesium hydroxide = 58.32 g/mole

First we have to calculate the moles of magnesium chloride.

`\text{ Moles of }MgCl_2=\frac{\text{ Mass of }MgCl_2}{\text{ Molar mass of }MgCl_2}=(4.6g)/(95.2g/mole)=0.048moles`

The given balanced reaction is,

`2NaOH(aq)+MgCl_2(aq)\rightarrow 2NaCl(aq)+Mg(OH)_2(s)`

From the reaction, we conclude that

1 mole of
`MgCl_2` react to give 1 mole of
`Mg(OH)_2`

0.048 moles of
`MgCl_2` react to give 0.048 moles of
`Mg(OH)_2`

Now we have to calculate the mass of
`Mg(OH)_2`

`\text{ Mass of }Mg(OH)_2=\text{ Moles of }Mg(OH)_2* \text{ Molar mass of }Mg(OH)_2`

`\text{ Mass of }Mg(OH)_2=0.048g* 58.32g/mole=2.799g`

The theoretical yield of magnesium hydroxide = 2.799 g

The experimental yield of magnesium hydroxide = 2.7 g

Now we have to calculate the percent yield.

Formula used :
`\%yield=\frac{\text{ Experimental yield}}{\text{ Theoretical yield}}* 100`

`\%yield=(2.7g)/(2.799g)* 100=96.46=96\%`

Therefore, the percent yield is, 96%