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Darya [45]
3 years ago
13

3cuso4 + 2al ------ 3cu + al2(so4)3 in the limiting reagent question above if you have 8.3 g of cuso4 and 20.1 g of al how many

grams of copper can you produce?
Chemistry
1 answer:
Alex17521 [72]3 years ago
8 0
Balanced equation for the reaction is;
3CuSO₄ + 2Al ---> Al₂(SO₄)₃ + 3Cu
Stoichiometry of CuSO₄ to Al is 3:2
Number of CuSO₄ moles present - 8.3 g/ 159.6 g/mol = 0.052 mol
Number of Al moles present - 20.1 g/ 27 g/mol = 0.74 mol
We need to first find the limiting reagent. Limiting reagent is the reactant that is fully consumed, amount of product formed depends on the amount of limiting reactant present.
3 mol of CuSO₄ reacts with 2 mol of Al
Therefore 0.052 mol of CuSO₄ reacts with - 2/3 x 0.052 = 0.034 mol of Al required.
But 0.74 mol of Al present , therefore Al is in excess and CuSO₄ is limiting reactant,
stoichiometry of CuSO₄ to Cu is 3:3 = 1:1
Therefore number of Copper moles formed - 0.052 mol
Then mass of Copper formed  - 0.052 mol x 63.5 g/mol = 3.30 g of Copper is produced

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What is the mass of 2.40 moles of magnesium chloride, mgcl2
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Hey there!

MgCl₂

Find molar mass of magnesium chloride.

Mg: 1 x 24.305

Cl: 2 x 35.453

--------------------  

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One mole of magnesium chloride has a mass of 95.211 grams.

We have 2.40 moles.

2.40 x 95.211 = 228.5

To 3 sig figs this is 229.

The mass of 2.40 moles of magnesium chloride is 229 grams.

Hope this helps!

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<h3>What is periodic table?</h3>

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4 0
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Hope this helps!
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