Answered

Get the answers you need at Westonci.ca, where our expert community is always ready to help with accurate information. Get immediate and reliable solutions to your questions from a knowledgeable community of professionals on our platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.

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

How Many Grams Of ZrCl4 Can Be Produced If 123 G Of ZrSiO4 React With 850 G Of Cl2 class=

Sagot :

mickymike92

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

Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.