Question

1. A solution contains an unknown amount of dissolved magnesium. Addition of0.0877 mol of Na2CO3 causes complete precipitation of all of the magnesium.What mass of magnesium was dissolved in the solution?

Answer 1

Answer:

4.25 g of magnesium.

Explanation:

What is given?

moles of Na2CO3 = 0.0877 moles.

molar mass of Mg (magnesium) = 24.3 g/mol.

Step-by-step solution:

First, let's state the chemical equation between Mg (magnesium) and Na2CO3:

$\text{2 Mg+Na}_2CO_3\rightarrow Mg_2CO_3+2Na.$

You can see that 1 mol of Na2CO3 reacts with 2 moles of Mg, so let's see how many moles of Mg are being produced by 0.0877 moles of Na2CO3:

$0.0877\text{ moles Na}_2CO_3\cdot\frac{2\text{ moles Mg}}{1\text{ mol Na}_2CO_3}=0.175\text{ moles Mg.}$

And the final step is to convert from 0.175 moles of Mg to grams using its molar mass. The conversion will look like this:

$0.175\text{ moles Mg}\cdot\frac{24.3\text{ g Mg}}{1\text{ mol Mg}}=4.25\text{ g Mg.}$

The answer is that there is 4.25 g of magnesium dissolved in the solution of 0.0877 moles of Na2CO.