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san4es73 [151]
3 years ago
12

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:

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

Keywords: excess reactant, limiting reactant

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If 10.0 grams of NaHCO3 is added to 10.0 g of HCl, determine the efficiency of baking soda as an antacid if 6.73 g of NaCl was p
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percentage yield of NaCl = 96.64%

Explanation:

The reaction was between NaHCO3 and HCl .The chemical equation can be represented below:

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The efficiency of NaHCO3 as an antacid , the limiting reactant is NaHCO3

as

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Therefore,

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since 84 g of NaHCO3 produces 58.5 g of NaCl

10 g of NaHCO3 will produce ? grams of NaCl

cross multiply

Theoretical yield of NaCl = (10 × 58.5)/84

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percentage yield of NaCl = actual yield/theoretical yield × 100

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percentage yield of NaCl = 673/6.9642857143

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