Question

3. The following reaction shows the synthesis of zinc citrate, Zn,(C6H5O7)2, an

ingredient in toothpaste, from zinc carbonate, ZnCO3, and citric acid, C,H,O7.
3ZnCO3(s) + 2C,H,O,(aq) → Zn3(C6H5O7)2(aq) + 3H₂O(l) + 3CO₂(g)
How many grams of citric acid would have to be used to produce 0.25mL of
water? (density of water = 0.997 g/mL)
a.

Answer 1

The balanced chemical equation for the reaction is: 3ZnCO3(s) + 2C6H8O7(aq) → Zn3(C6H5O7)2(aq) + 3H2O(l) + 3CO2(g)

From the equation, we can see that for every 3 moles of water produced, 2 moles of citric acid are consumed. The molar mass of citric acid (C6H8O7) is 192.12 g/mol.

0.25 mL of water has a mass of 0.25 mL * 0.997 g/mL = 0.24925 g The number of moles of water produced is 0.24925 g / 18.02 g/mol = 0.01383 mol The number of moles of citric acid consumed is (2/3) * 0.01383 mol = 0.00922 mol The mass of citric acid consumed is 0.00922 mol * 192.12 g/mol = 1.77 g

So, to produce 0.25 mL of water, 1.77 grams of citric acid would have to be used.