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

Chemistry

2 answers:

KiRa [710]3 years ago

6 0

Answer:

14.7 g

Explanation:

Mrac [35]3 years ago

4 0

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:

Calculate the moles of product (Cu) formed.
Use the stoichiometric ratio of the product (Cu) and the limiting reactant (Cr) to get the number of moles of Cr used up.
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

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