# What is the theoretical mass of copper (ii) carbonate if 0.565 g of copper (ii) sulfate reacts with 0.465 g of sodium carbonate?

Oct 3, 2016

The theoretical mass of ${\text{CuCO}}_{3}$ is 0.437 g.

#### Explanation:

This is a limiting reactant problem.

We know that we will need a balanced equation, molar masses, and moles of the compounds involved.

1. Gather all the information in one place with the molecular masses above the formulas and the given masses and moles below them.

${M}_{r} : \textcolor{w h i t e}{m m m m} 159.61 \textcolor{w h i t e}{m m} 105.99 \textcolor{w h i t e}{m m l l} 123.55$
$\textcolor{w h i t e}{m m m m m m} {\text{CuSO"_4 + "Na"_2"CO"_3 → "CuCO"_3 + "Na"_2"SO}}_{4}$
$\text{Mass/g:} \textcolor{w h i t e}{m m l l} 0.565 \textcolor{w h i t e}{m m l l} 0.465$
$\text{Amt/mol:"color(white)(ll)"0.003 540"color(white)(ll)"0.004 387}$
$\text{Divide by:} \textcolor{w h i t e}{m m m} 1 \textcolor{w h i t e}{m m m m l} 1$
$\text{Moles rxn:"color(white)(ll)"0.003 540"color(white)(ll)"0.004 387}$

2. Identify the limiting reactant

An easy way to identify the limiting reactant is to calculate the "moles of reaction" each will give:

You divide the moles of each reactant by its corresponding coefficient in the balanced equation.

I did that for you in the table above.

${\text{CuSO}}_{4}$ is the limiting reactant because it gives fewer moles of reaction.

3. Calculate the mass of ${\text{CuCO}}_{3}$

${\text{Mass of CuCO"_3 = "0.003 540" color(red)(cancel(color(black)("mol CuSO"_4))) × (1 color(red)(cancel(color(black)("mol CuCO"_3))))/(1 color(red)(cancel(color(black)("mol CuSO"_4)))) × ("123.55 g CuCO"_3)/(1 color(red)(cancel(color(black)("mol CuCO"_3)))) = "0.437 g CuCO}}_{3}$