Question

To make a phosphorus fertilizer, agricultural companies use the following reaction:

Ca3P₂O8 + 2H₂SO4 + 4H₂O → CaH₂P₂O + 2aH4SO
fertilizer
If 9.80 x 10³ grams of H₂SO4 are reacted with excess Ca,P₂O, and H₂O, how many grams
of fertilizer can be made?

Answer 1

To determine the number of grams of fertilizer that can be made, we need to calculate the stoichiometric ratio between H₂SO₄ and fertilizer in the given balanced chemical equation.

From the balanced equation:
1 mole of Ca₃P₂O₈ + 2 moles of H₂SO₄ + 4 moles of H₂O → 1 mole of CaH₂P₂O₆ + 2 moles of H₄SO₆ (fertilizer)

First, we need to convert the given mass of H₂SO₄ to moles:
9.80 x 10³ grams H₂SO₄ x (1 mole H₂SO₄ / molar mass of H₂SO₄)

The molar mass of H₂SO₄ is:
2(1.01 g/mol of H) + 32.07 g/mol of S + 4(16.00 g/mol of O) = 98.09 g/mol

Calculating the moles of H₂SO₄:
9.80 x 10³ g H₂SO₄ / 98.09 g/mol ≈ 100 moles H₂SO₄

Based on the stoichiometry of the balanced equation, we can determine the moles of fertilizer produced:
100 moles H₂SO₄ x (1 mole fertilizer / 2 moles H₂SO₄)

The ratio is 1 mole of fertilizer for every 2 moles of H₂SO₄.

Calculating the moles of fertilizer:
100 moles H₂SO₄ / 2 = 50 moles of fertilizer

Finally, we can convert moles of fertilizer to grams using the molar mass of the fertilizer compound. Since the molar mass of the fertilizer compound (CaH₂P₂O₆) is not provided, we cannot provide the exact mass of the fertilizer that can be made.