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Impure phosphoric acid for use in the manufacture of fertilizers is produced by the reaction of sulfuric acid on phosphate rock, of which a principal component is Ca3(PO4)2. The reaction is Ca3(PO4)2(s) + 3 H2SO4(aq) → 3 CaSO4(s) + 2 H3PO4(aq) How many moles of H3PO4 can be produced from the reaction of 186.8 kg of H2SO4? Answer in units of mol.

Respuesta :

IthaloAbreu

Answer:

1271 mol

Explanation:

The balanced reaction is

Ca₃(PO₄)₂(s) +3H₂SO₄(aq) →  3CaSO₄(s) + 2H₃PO₄(aq)

So, the stoichiometry of the reaction (the proportion of the substances) is 3 mol of H₂SO₄ for 2 mol of H₃PO₄. The molar masses are H= 1 g/mol, S = 32 g/mol, O = 16 g/mol, so:

H₂SO₄  = 2x1 + 1x32 + 4x16 = 98 g/mol

The number of moles is the mass divided by the molar mass, so for 186.8 kg (186800 g) of H₂SO₄

n = 186800/98 = 1906.12 mol

So, the stoichiometry calculus is:

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

1906.12 mol of H₂SO₄ ------------ x

By a simple direct three rule:

3x = 3812.24

x = 1271 mol of H₃PO₄

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