Answer:
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
