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Which equation shows how to calculate how many grams (g) of \( \text{KOH} \) would be needed to fully react with \( 4 \, \text{mol} \, \text{Mg}(\text{OH})_2 \)? The balanced reaction is:

[tex]\[ \text{MgCl}_2 + 2 \, \text{KOH} \rightarrow \text{Mg}(\text{OH})_2 + 2 \, \text{KCl} \][/tex]

A. \(\frac{4 \, \text{mol} \, \text{Mg}(\text{OH})_2}{1} \times \frac{2 \, \text{mol} \, \text{KOH}}{1 \, \text{mol} \, \text{Mg}(\text{OH})_2} \times \frac{56.10 \, \text{g} \, \text{KOH}}{1 \, \text{mol} \, \text{KOH}}\)

B. \(\frac{4 \, \text{mol} \, \text{Mg}(\text{OH})_2}{1} \times \frac{1 \, \text{mol} \, \text{KOH}}{2 \, \text{mol} \, \text{Mg}(\text{OH})_2} \times \frac{56.10 \, \text{g} \, \text{KOH}}{1 \, \text{mol} \, \text{KOH}}\)

C. \(\frac{1 \, \text{mol} \, \text{MgCl}_2}{1} \times \frac{2 \, \text{mol} \, \text{KOH}}{1 \, \text{mol} \, \text{MgCl}_2} \times \frac{56.10 \, \text{g} \, \text{KOH}}{1 \, \text{mol} \, \text{KOH}}\)

D. [tex]\(\frac{1 \, \text{mol} \, \text{Mg}(\text{OH})_2}{1} \times \frac{2 \, \text{mol} \, \text{KOH}}{1 \, \text{mol} \, \text{Mg}(\text{OH})_2} \times \frac{56.10 \, \text{g} \, \text{KOH}}{1 \, \text{mol} \, \text{KOH}}\)[/tex]

Sagot :

GinnyAnsweridn
To determine how many grams of \( KOH \) are needed to fully react with \( 4 \, \text{mol} \, Mg(OH)_2 \) according to the balanced reaction:

[tex]\[ MgCl_2 + 2 KOH \rightarrow Mg(OH)_2 + 2 KCl \][/tex]

we need to go through the stoichiometric calculations step-by-step.

First, we need to identify the molar relationship given in the balanced equation:
[tex]\[ 2 \, \text{mol} \, KOH \rightarrow 1 \, \text{mol} \, Mg(OH)_2 \][/tex]

We are given \( 4 \, \text{mol} \, Mg(OH)_2 \). Using stoichiometry:
[tex]\[ 4 \, \text{mol} \, Mg(OH)_2 \times \frac{2 \, \text{mol} \, KOH}{1 \, \text{mol} \, Mg(OH)_2} \][/tex]

Next, we calculate the amount of \( KOH \) needed in moles:
[tex]\[ 4 \times 2 = 8 \, \text{mol} \, KOH \][/tex]

Now, we convert the moles of \( KOH \) to grams. The molar mass of \( KOH \) is given as \( 56.10 \, \text{g/mol} \). Using this, we perform the conversion:
[tex]\[ 8 \, \text{mol} \, KOH \times 56.10 \, \text{g/mol} \, KOH \][/tex]

This calculation gives:
[tex]\[ 8 \times 56.10 = 448.8 \, \text{grams} \, KOH \][/tex]

Therefore, the total grams of \( KOH \) needed to fully react with \( 4 \, \text{mol} \, Mg(OH)_2 \) is \( 448.8 \, \text{grams} \).

Upon analyzing the provided options, we find that:

- Option A is the correct equation:
[tex]\[ \frac{4 \, \text{mol} \, Mg(OH)_2}{1} \times \frac{2 \, \text{mol} \, KOH}{1 \, \text{mol} \, Mg(OH)_2} \times \frac{56.10 \, \text{g} \, KOH}{1 \, \text{mol} \, KOH} \][/tex]
This matches our detailed, step-by-step calculation. Thus, [tex]\( 4 \, \text{mol} \, Mg(OH)_2 \)[/tex] needs [tex]\( 448.8 \, \text{g} \, KOH \)[/tex] to fully react.
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