Considering the reaction stoichiometry, the correct answer is option C: the equation \(4 moles KOHx\frac{1 mol Mg(OH)_{2} }{2 moles KOH} x\frac{58.31 g Mg(OH)_{2} }{1 mol Mg(OH)_{2} }\) shows how to calculate how many grams (g) of Mg(OH)₂ would be produced from 4 mol KOH.
The balanced reaction is:
MgCl₂ + 2 KOH → Mg(OH)₂ + 2 KCl
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- MgCl₂: 1 mole
- KOH: 2 moles
- Mg(OH)₂: 1 mole
- KCl: 2 moles
On the other side, you know that the molar mass of Mg(OH)₂ is 58.31 \(\frac{g}{mole}\), this is the amount of mass that a substance contains in one mole.
Then, knowing that by stoichiometry
2 moles of KOH are produced together with 1 mole of Mg(OH)₂, considering the molar mass of Mg(OH)₂, the expression to calculate the mass of Mg(OH)₂ that is produced together with 4 moles of KOH is :
\(4 moles KOHx\frac{1 mol Mg(OH)_{2} }{2 moles KOH} x\frac{58.31 g Mg(OH)_{2} }{1 mol Mg(OH)_{2} }\)
In summary, the correct answer is option C: the equation \(4 moles KOHx\frac{1 mol Mg(OH)_{2} }{2 moles KOH} x\frac{58.31 g Mg(OH)_{2} }{1 mol Mg(OH)_{2} }\) shows how to calculate how many grams (g) of Mg(OH)₂ would be produced from 4 mol KOH.
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