Question

How many grams of sodium carbonate are produced when 5.3 moles of sodium phosphate reacts with aluminum carbonate?

Answer 1

Answer: There is 842.54 grams of sodium carbonate are produced when 5.3 moles of sodium phosphate reacts with aluminum carbonate.

Explanation:

Chemical equation depicting reaction between sodium phosphate and aluminum carbonate is as follows.

$Al_{2}(CO_{3})_{3} + 2Na_{3}PO_{4} \rightarrow 2AlPO_{4} + 3Na_{2}CO_{3}$

As this equation contains same number of atoms on both reactant and product side. So, this equation is a balanced equation.

According to the equation, 2 moles of sodium phosphate is giving 3 moles of sodium carbonate.

Therefore, sodium carbonate formed by 5.3 moles of sodium phosphate is as follows.

$\frac{3}{2} \times 5.3 mol\\= 7.95 mol$

As number of moles is the mass of substance divided by its molar mass. So, mass of sodium carbonate ( molar mass = 105.98 g/mol) is as follows.

$No. of moles = \frac{mass}{molar mass}\\7.95 mol = \frac{mass}{105.98 g/mol}\\mass = 842.54 g$

Thus, we can conclude that there is 842.54 grams of sodium carbonate are produced when 5.3 moles of sodium phosphate reacts with aluminum carbonate.