Question

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

Answer 1

Taking into account the reaction stoichiometry, 139.68 grams of ZrCl₄ are formed when 123 g of ZrSiO₄ react with 85.0 g of Cl₂.

Reaction stoichiometry

In first place, the balanced reaction is:

ZrSiO₄ + 2 Cl₂ → ZrCl₄ + SiO₂ + O₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

• ZrSiO₄: 1 mole
• Cl₂: 2 moles
• ZrCl₄: 1 mole
• SiO₂: 1 mole
• O₂: 1 mole

The molar mass of the compounds is:

• ZrSiO₄: 183.22 g/mole
• Cl₂: 70.9 g/mole
• ZrCl₄: 233.02 g/mole
• SiO₂: 60 g/mole
• O₂: 32 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

• ZrSiO₄: 1 mole ×183.22 g/mole= 183.22 grams
• Cl₂: 2 moles ×70.9 g/mole= 141.8 grams
• ZrCl₄: 1 mole ×233.02 g/mole= 233.02 grams
• SiO₂: 1 mole ×60 g/mole= 60 grams
• O₂: 1 mole ×32 g/mole= 32 grams

Limiting reagent

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 183.22 grams of ZrSiO₄ reacts with 141.8 grams of Cl₂, 123 grams of ZrSiO₄ reacts with how much mass of Cl₂?

mass of Cl₂= (123 grams of ZrSiO₄×141.8 grams of Cl₂)÷ 183.22 grams of ZrSiO₄

mass of Cl₂= 95.19 grams

But 95.19 grams of Cl₂ are not available, 85 grams are available. Since you have less mass than you need to react with 123 grams of ZrSiO₄, Cl₂ will be the limiting reagent.

Mass of ZrCl₄ formed

The following rules of three can be applied: if by reaction stoichiometry 141.8 grams of Cl₂ form 233.02 grams of ZrCl₄, 85 grams of Cl₂ form how much mass of ZrCl₄?

mass of ZrCl₄= (85 grams of Cl₂× 233.02 grams of ZrCl₄)÷ 141.8 grams of Cl₂

mass of ZrCl₄= 139.68 grams

Finally, 139.68 grams of ZrCl₄ are formed when 123 g of ZrSiO₄ react with 85.0 g of Cl₂.

Answer 2

153.3 grams of ZrCl₄ are produced

Step-by-step 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