Question

Which equation shows how to calculate how many grams (g) of Mg(OH)2 would be produced from 4 mol KOH? The balanced reaction is: MgCl2 + 2KOH -> Mg(OH)2 + 2KCl

Answer 1

The answer is C.

Since KOH and MgCl2  are in a 2:1 ratio, the moles of MgCl2 formed will always be half of the moles of KOH. The only answer where moles of KOH are halved is option C

Answer 2

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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