Question

Using the following equation: 3 Mg(OH)2 + 2 H3PO4 --> Mg3(PO4)2 + 6 H2OIf you start with 17 grams of H3PO4, how many moles of Magnesium phosphate will you produce?

Answer 1

Step-by-step explanation:

Magnesium hydroxide and phosphoric acid will react according to the following reaction.

3 Mg(OH)₂ + 2 H₃PO₄ --> Mg₃(PO₄)₂ + 6 H₂O

We have to find the number of moles of Mg₃(PO₄)₂ that can be produced from 17 g of H₃PO₄. First we have to convert the 17 g into moles using the molar mass of H₃PO₄.

molar mass of H₃PO₄ = 97.99 g/mol

moles of H₃PO₄ = 17 g * 1 mol/(97.99 g)

moles of H₃PO₄ = 0.173 moles

3 Mg(OH)₂ + 2 H₃PO₄ --> Mg₃(PO₄)₂ + 6 H₂O

Now, if we pay attention to the coefficients of the equation we see that it says: 3 moles of Mg(OH)₂ will react with 2 moles of H₃PO₄ to give 1 mol of Mg₃(PO₄)₂ and 6 moles of H₂O. So the molar ratio between H₃PO₄ and Mg₃(PO₄)₂ is 2 to 1. We can use that relationship to find the number of moles of Mg₃(PO₄)₂ that can be produced.

2 moles of H₃PO₄ : 1 mol of Mg₃(PO₄)₂ molar ratio

moles of Mg₃(PO₄)₂ = 0.173 moles of H₃PO₄ * 1 mol of Mg₃(PO₄)₂/(2 moles of H₃PO₄)

moles of Mg₃(PO₄)₂ = 0.0867 moles

Answer: we will produce 0.0867 moles of Mg₃(PO₄)₂ if we start with 17 g of H₃PO₄.