select the correct answer.\nexcess sodium hydroxide is added to a solution containing 4.6 grams of magnesium…

Explanation:

Step1: Calculate molar mass of MgCl₂

The molar mass of Mg (magnesium) is approximately 24.31 g/mol and Cl (chlorine) is approximately 35.45 g/mol. For MgCl₂, molar mass $M_{MgCl_2}=24.31 + 2\times35.45=95.21$ g/mol.

Step2: Calculate moles of MgCl₂

Given mass of MgCl₂ is $m = 4.6$ g. Moles of MgCl₂, $n_{MgCl_2}=\frac{m}{M_{MgCl_2}}=\frac{4.6}{95.21}\approx0.0483$ mol.

Step3: Determine moles of Mg(OH)₂ produced theoretically

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

Step4: Calculate molar mass of Mg(OH)₂

The molar mass of Mg is 24.31 g/mol, O is 16 g/mol and H is 1 g/mol. For Mg(OH)₂, molar mass $M_{Mg(OH)_2}=24.31+2\times(16 + 1)=58.31$ g/mol.

Step5: Calculate theoretical mass of Mg(OH)₂

Theoretical mass of Mg(OH)₂, $m_{theo}=n_{theo}\times M_{Mg(OH)_2}=0.0483\times58.31\approx2.82$ g.

Step6: Calculate percent yield

Percent yield $=\frac{\text{actual yield}}{\text{theoretical yield}}\times100%=\frac{2.7}{2.82}\times100%\approx96%$.

Answer:

D. 96%