How many grams of cobalt(II) chloride are needed to react completely with 68.9 mL of .366 M KOH solution, if the equation for the reaction is Co^(2+) (aq) + 2OH^(-)(aq) -> Co(OH)_2(s)?

Nov 27, 2015

$\text{1.64 g}$

Explanation:

Your strategy for this problem will be to use the molarity and volume of the potassium hydroxide solution, $\text{KOH}$, to find the number of moles of potassium hydroxide that take part in the reaction.

Once you know how many moles of potassium hydroxide you have, you can use the mole ratio that exists between the two compounds to determine the number of moles of cobalt(II) chloride needed for the reaction.

So, molarity is defined as moles of solute divided by liters of solution.

$\textcolor{b l u e}{c = \frac{n}{V}}$

This means that the number of moles of potassium hydroxide will be equal to

$\textcolor{b l u e}{c = \frac{n}{V} \implies n = c \cdot V}$

${n}_{K O H} = \text{0.366 M" * 68.9 * 10^(-3)"L" = "0.02522 moles KOH}$

Use the $1 : 2$ mole ratio that exists between the two compounds to get the number of moles of cobalt(II) chloride needed for the reaction

0.02522 color(red)(cancel(color(black)("moles KOH"))) * ("1 mole CoCl"_2)/(2color(red)(cancel(color(black)("moles KOH")))) = "0.01261 moles CoCl"_2

Finally, use cobalt(II) chloride's molar mass to determine how many grams would contain this many moles

0.01261 color(red)(cancel(color(black)("moles"))) * "129.839 g"/(1color(red)(cancel(color(black)("mole")))) = color(green)("1.64 g CoCl"_2)