How many grams of ZrCl4 can be produced if 123 g of ZrSiO4 react with 85.0 g of Cl2?

153.3 grams of ZrCl₄ are produced

Explanation:

The equation of the reaction is as follows:

ZrSiO₄ + Cl₂ > ZrCl₄ + SiO₂ + O₂

molar mass of ZrSiO₄ = (91 + 32 + 16 * 4) = 187.0 g/mol

molar mass of ZrCl₄ = (91 + 35.5 * 4) = 233.0 g/mol

molar mass of Cl₂ = (35.5 * 2) 71.0 g/mol

From the equation of reaction, 1 mole of ZrSiO₄ reacts with one mole of  Cl₂ to produce one mole of ZrCl₄

number of moles of ZrSiO₄ present in 123 g = 123/187 = 0.65 moles

number of moles of Cl₂ present in 85.0 g = 85.0/71.0 = 1.19 moles

therefore, ZrSiO₄ is the limiting reactant

123.0 g of ZrSiO₄ will react with excess Cl₂ to produce 123 * 233/187 grams of ZrCl₄ = 153.3 grams of ZrCl₄

Therefore, 153.3 grams of ZrCl₄ are produced