Respuesta :
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.
[tex]\text{ Moles of }MgCl_2=\frac{\text{ Mass of }MgCl_2}{\text{ Molar mass of }MgCl_2}=\frac{4.6g}{95.2g/mole}=0.048moles[/tex]
The given balanced reaction is,
[tex]2NaOH(aq)+MgCl_2(aq)\rightarrow 2NaCl(aq)+Mg(OH)_2(s)[/tex]
From the reaction, we conclude that
1 mole of [tex]MgCl_2[/tex] react to give 1 mole of [tex]Mg(OH)_2[/tex]
0.048 moles of [tex]MgCl_2[/tex] react to give 0.048 moles of [tex]Mg(OH)_2[/tex]
Now we have to calculate the mass of [tex]Mg(OH)_2[/tex]
[tex]\text{ Mass of }Mg(OH)_2=\text{ Moles of }Mg(OH)_2\times \text{ Molar mass of }Mg(OH)_2[/tex]
[tex]\text{ Mass of }Mg(OH)_2=0.048g\times 58.32g/mole=2.799g[/tex]
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 : [tex]\%yield=\frac{\text{ Experimental yield}}{\text{ Theoretical yield}}\times 100[/tex]
[tex]\%yield=\frac{2.7g}{2.799g}\times 100=96.46=96\%[/tex]
Therefore, the percent yield is, 96%
