If 12.85 g of chromium metal is reacted with 10.72 g of phosphoric acid, then what is the maximum mass in grams of chromium(iii) phosphate that can be produced? 2cr(s) + 2h3po4(aq) → 2crpo4(s) + 3h2(g)

16.02 g

We are given;

We are require to calculate the maximum mass of chromium (III) phosphate that can be produced.

2Cr(s) + 2H₃PO₄(aq) → 2CrPO₄(s) + 3H₂(g)

Moles = Mass ÷ Molar mass

Molar mass of chromium = 52.0 g/mol

Moles of Chromium = 12.85 g ÷ 52.0 g/mol

                                 = 0.247 moles

Molar mass of phosphoric acid = 97.994 g/mol

Moles of phosphoric acid = 10.72 g ÷ 97.994 g/mol

                                          = 0.109 moles

Hence; moles of CrPO₄ = 0.109 moles

We know that, mass = Moles × Molar mass

Molar mass of  CrPO₄ = 146.97 g/mol

Thus,

Mass of  CrPO₄ = 0.109 moles × 146.97 g/mol

                         = 16.02 g

Thus, the maximum mass of CrPO₄ that can be produced is 16.02 g

o conteúdo de calor de um sistema é igual à entalpia apenas para um sistema que é constante