Type the correct answer in the box. Express your answer to two significant figures.
You are performing a reaction with 1.7 moles of hydroiodic acid and 3.43 moles of zinc bromide:
[tex]2 HI+ZnBr_2 \rightarrow 2 HBr+ZnI_2[/tex]
How many moles of zinc iodide can be made?
The theoretical yield is $\square$ moles of zinc iodide.
Answer (2)
Determine the limiting reactant: HI is the limiting reactant because the mole ratio of HI to Z n B r 2 is 2:1, and we have 1.7 moles of HI and 3.43 moles of Z n B r 2 .
Calculate the moles of Z n B r 2 required to react completely with 1.7 moles of HI: 1.7 m o l es ( H I ) × ( 1 m o l e Z n B r 2 /2 m o l es H I ) = 0.85 m o l es Z n B r 2 .
Calculate the moles of Z n I 2 produced from 1.7 moles of HI: 1.7 m o l es ( H I ) × ( 1 m o l e Z n I 2 /2 m o l es H I ) = 0.85 m o l es Z n I 2 .
The theoretical yield of zinc iodide is 0.85 moles.
Explanation
Problem Analysis We are given the balanced chemical equation: 2 H I + Z n B r 2 → 2 H B r + Z n I 2 We have 1.7 moles of hydroiodic acid (HI) and 3.43 moles of zinc bromide ( Z n B r 2 ). Our goal is to find the theoretical yield of zinc iodide ( Z n I 2 ) in moles.
Identifying the Limiting Reactant To determine the theoretical yield, we first need to identify the limiting reactant. The limiting reactant is the reactant that is completely consumed in the reaction, and it determines the maximum amount of product that can be formed. From the balanced equation, the stoichiometric ratio of HI to Z n B r 2 is 2:1.
Calculating Moles of Zinc Bromide Required Let's calculate how many moles of Z n B r 2 are required to react completely with 1.7 moles of HI: m o l es ( Z n B r 2 ) = 1.7 m o l es ( H I ) × 2 m o l es H I 1 m o l e Z n B r 2 = 0.85 m o l es Z n B r 2 Since we have 3.43 moles of Z n B r 2 , which is more than the 0.85 moles required, HI is the limiting reactant.
Calculating Moles of Zinc Iodide Produced Now we calculate the moles of Z n I 2 produced from 1.7 moles of HI using the stoichiometric ratio: m o l es ( Z n I 2 ) = 1.7 m o l es ( H I ) × 2 m o l es H I 1 m o l e Z n I 2 = 0.85 m o l es Z n I 2
Final Answer The theoretical yield of zinc iodide is 0.85 moles. Since we need to express the answer to two significant figures, the answer remains 0.85 moles.
Examples
In chemical manufacturing, determining the limiting reactant and theoretical yield is crucial for optimizing production. For instance, if a pharmaceutical company is synthesizing a drug, knowing the exact amount of each reactant needed and the maximum possible yield helps minimize waste and maximize profit. This ensures efficient use of resources and accurate cost estimation, which are vital for large-scale production.
The theoretical yield of zinc iodide (ZnI₂) that can be produced from 1.7 moles of hydroiodic acid (HI) is 0.85 moles. Hydroiodic acid is the limiting reactant because it will be completely consumed in the reaction. The answer is expressed to two significant figures, which is 0.85 moles.
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