# How many grams of K_2CrO_4 are formed per gram of K_2CO_3 used?

Aug 5, 2018

For every gram of $\text{K"_2"CO"_3}$ used, $\text{2.82 g K"_2"CrO"_4}$ is formed.

#### Explanation:

Balanced equation

$\text{K"_2"Cr"_2"O"_7 + "K"_2"CO"_3}$$\rightarrow$$\text{2K"_2"CrO"_4 + "CO"_2}$

Determine mol $\text{K"_2"CO"_3}$ are in $\text{1.00 g}$. ($\text{1.00 g}$ is arbitrary, since the number of significant figures were not specified. You can redo the calculations with the number of significant figures you wish.) Divide the mass of $\text{K"_2"CO"_3}$ by its molar mass $\left(\text{138.205 g/mol}\right)$ by multiplying the mass of $\text{K"_2"CO"_3}$ by the inverse of its molar mass $\left(\text{1 mol"/"138.205 g}\right)$.

1.00color(red)cancel(color(black)("g K"_2"CO"_3))xx(1"mol K"_2"CO"_3)/(138.205color(red)cancel(color(black)("g K"_2"CO"_3)))="0.007236 mol K"_2"CO"_3"

I am keeping an extra digit to reduce rounding errors. I will round to three significant figures at the end.

Determine mol $\text{K"_2"CrO"_4}$ by multiplying mol $\text{K"_2"CO"_3}$ by the mol ratio between the two compounds in the balanced equation, with mol $\text{K"_2"CrO"_4}$ in the numerator.

0.007236color(red)cancel(color(black)("mol K"_2"CO"_3))xx(2"mol K"_2"CrO"_4)/(1color(red)cancel(color(black)("mol K"_2"CO"_3)))="0.01447 mol K"_2"CrO"_4"

Determine mass $\text{K"_2"CrO"_4}$ by multiplying mol $\text{K"_2"CrO"_4}$ by its molar mass $\left(\text{194.189 g/mol}\right)$.

0.01447color(red)cancel(color(black)("mol K"_2"CrO"_4))xx(194.189"g K"_2"CrO"_4)/(1color(red)cancel(color(black)("mol K"_2"CrO"_4)))="2.81 g K"_2"CrO"_4" (rounded to three significant figures)

For every gram of $\text{K"_2"CO"_3}$ used, $\text{2.81 g K"_2"CrO"_4}$ is formed.