# How many grams of potassium sulfide are produced from the reaction of 1.23 moles of potassium...

## Question:

How many grams of potassium sulfide are produced from the reaction of 1.23 *moles* of potassium with sulfur?

## Molar Ratio:

The balanced chemical reaction is a way to determine the molar ratio between substances in a given reaction. The molar ratio is obtained by taking the ratio between the coefficients of the concerned molecules in the given balanced chemical reaction.

## Answer and Explanation:

Determine the mass of the product, {eq}\displaystyle m_{K_2S} {/eq}, by multiplying the given number of moles of {eq}\displaystyle K {/eq}, {eq}\displaystyle n_{K} = 1.23\ mol\ K {/eq}, to the molar ratio between {eq}\displaystyle K_2S {/eq} and {eq}\displaystyle K {/eq} in the balanced chemical reaction, and finally multiplying the molar mass of {eq}\displaystyle K_2S {/eq} (MW = 110.262 g/mol). The reaction of the substances is given as {eq}\displaystyle 2\ K + S \to K_2S {/eq}, so we know that there are {eq}\displaystyle 2\ mol\ K {/eq} to produce {eq}\displaystyle 1\ mol\ K_2S {/eq}. We proceed with the solution.

{eq}\begin{align} \displaystyle m_{K_2S} &= 1.23\ mol\ K\times \frac{1\ mol\ K_2S}{2\ mol\ K}\times 110.262\ \rm{g/mol}\\ &= 67.8\ g\ K_2S \end{align} {/eq}

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from Chemistry 101: General Chemistry

Chapter 9 / Lesson 2