Question

In the following reaction 108 grams of water were formed

3Cu(OH)₂ + 2H₃ PO₄ -----> Cu₃ (PO₄)₂ + 6H₂O

A) Calculated how many grams of copper(II) hydroxide were reacted?

B) Calculated how many grams of phoosphoric acid H₃PO₄ were reacted?

C) Calculated how many grams of copper (II) phosphate were formed?

Answer 1

3Cu(OH)2 + 2H3 PO4 -----> Cu3 (PO4)2 + 6H2O

Three moles of Cu(OH)2 reacts with two moles of H3 PO4 to produce one mole of Cu3 (PO4)2 and moles of H2O

Amount of water produced: 108 g

Molecular weight of water: 18g/mol

Number of moles of water produced = 180g/ 18 gmol^-1 = 6 moles

Since 6 moles are produces , so the stoichiomety of the reaction is as shown above.

A) Calculated how many grams of copper(II) hydroxide were reacted?

Molecular weight of Ca(OH)2: 74.093 g/mol

So three moles of Ca(OH)2 have reacted = 74.093 g/mol x 3 moles = 222.279 g

B) Calculated how many grams of phoosphoric acid H3PO4 were reacted?

Molecular weight of H3PO4: 97.994 g/mol

So two moles of H3PO4 have reacted = 97.994 g/mol x 2 moles = 195.998 g

C) Calculated how many grams of copper (II) phosphate were formed?

Molecular weight of Cu3(PO4)2: 380.58 g/mol

So one mole of Cu3(PO4)2 is produced =380.58 g/mol x 1 moles = 380.58 g