Question

How many grams of chromium are needed to react with an excess of CuSO4 to produce 27.0g Cu

Answer 1

14.7 g of chromium are need to produce 27.0 g of copper in the reaction:

2 Cr + 3 CuSO4 → Cr2(SO4)3 + 3 Cu. In this reaction, chromium is the limiting reactant and CuSO4 is the excess reactant.

Further Explanation:

The problem given is an example of a stoichiometry problem. Stoichiometry involves the determination of the amounts of products formed and amount of reactants consumed in a chemical reaction. It uses the ratio of the reactants and products given by the balanced chemical equation.

Limiting reactant is the reactant that determines how much of the product(s) will be obtained. In this problem, chromium is the limiting reactant.

To solve the amount of limiting reactant we use several steps:

1. Calculate the moles of product (Cu) formed.
2. Use the stoichiometric ratio of the product (Cu) and the limiting reactant (Cr) to get the number of moles of Cr used up.
3. Convert the moles of Cr used up to mass (in grams).

STEP 1: Convert mass to moles using the equation below:

$no. \ of \ moles \ = given \ mass \ (\frac{1 \ mole}{molar \ mass})$

For this problem,

$no. \ of \ moles \ Cu \ = 27.0 \ g \ Cu \ (\frac{1 \ mole \ Cu}{ 63.55 \ g}) \\\boxed {no. \ of \ moles \ Cu \ = 0.4249}$

STEP 2: Calculate the number of moles of Cr used up using the stoichiometric ratio from the balanced equation: 3 moles Cu: 2 moles Cr

$moles \ of \ Cr \ = given \ moles \ Cu (\frac{2 \ mol \ Cr}{3 \ mol \ Cu})\\moles \ of \ Cr \ = \ 0.4249\ mol\ Cu (\frac{2 \ mol \ Cr}{3 \ mol \ Cu})\\ \\\boxed {moles \ of \ Cr \ = \ 0.2832 }$

STEP 3: Convert moles of Cr to mass (in grams).

$mass \ = given \ moles \ (\frac{molar \ mass}{1 \ mole})$

To get the mass of Cr,

$mass \ of \ Cr \ = 0.2832 \ mol \ (\frac{52.00 \ g}{1 \ mol})\\\boxed {mass \ of \ Cr \ = 14.7 \ g}$

Learn More

1. Learn more about stoichiometry https://brainly.com/question/11292649
2. Learn more about limiting reactant https://brainly.com/question/7144022
3. Learn more about moles https://brainly.com/question/2293005

Keywords: excess reactant, limiting reactant

Answer 2

The mass of chromium, Cr needed to react with excess CuSO₄ to produce 27 g of Cu is 14.7 g

Balanced equation

2Cr + 3CuSO₄ → Cr₂(SO₄)₃ + 3Cu

Molar mass of Cu = 52 g/mole

Mass of Cr from the balanced equation = 2 × 52 = 104 g

Molar mass of Cu = 63.5 g/mole

Mass of Cu from the balanced equation = 3 × 63.5 = 190.5 g

SUMMARY

From the balanced equation above,

190.5 g of Cu were produced from 104 g of Cr

How to determine the mass of Cr needed

From the balanced equation above,

190.5 g of Cu were produced from 104 g of Cr

Therefore,

27 g of Cu will be produce by = (27 × 104) / 190.5 = 14.7 g of Cr

Thus, 14.7 g of Cr is needed for the reaction

Learn more about stoichiometry:

https://brainly.com/question/14735801

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