Question

Impure phosphoric acid for use the manufacture of fertilizers is produced by the reaction of sulfuric acid on phosphate rock, of which a principle component is Ca3(Po4)2. The reaction is Ca3(Po4)2 + 3H2SO4 -> 3 CaSO4 + 2H3PO4How many molds of H3PO4 can be produced from the reaction of 180.4 Kg of H2SO4? Answer in units of moles

Answer 1

The question asks us to calculate how many moles of phosporic acid (H3PO4) can be produced from 180.4 kg of sulfuric acid (H2SO4).

The first step is to determine the molar mass of each compound. To calculate the molar mass, I'll be using the following atomic masses (found in a periodic table):

H: 1.00 u

P: 30.97 u

O: 15.99 u

S: 32,07 u

Then we calculate the molar mass of H3PO4 and H2SO4 considering the atomic masses and the amount of each atom in the compound:

H3PO4: (3 * 1.00) + (1 * 30.97) + (4 * 15.99) = 97.93 g/mol

H2SO4: (2 * 1.00) + (1 * 32.07) + (4 * 15.99) = 98.03 g/mol

Now that we know the molar mass, we can calculate the number of moles contained in 180.4 kg of H2SO4::

98.03 g H2SO4 ------------- 1 mol H2SO4

(180.4 * 1000) g H2SO4 -- x

Solving for x, we have that 180.4 kg of H2SO4 corresponds to 1.840 x 10^3 moles of H2SO4

The last step is to calcuate the amount of H3PO4 produced considering the stoichiometric coefficients from the reaction given:

3 mol H2SO4 -------------------- 2 mol H3PO4

1.840 x 10^3 mol H2SO4 ---- y

Solving for y, we have that 1.227 x 10^3 moles of H3PO4 are produced when using 180.4 kg of H2SO4.