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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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4 0
3 years ago
C5H12 + 8O2 → 5CO2 + 6H2O<br><br> Calculate the energy change for the reaction:
KATRIN_1 [288]
C5H12 (l) + 8O2 (g) ----> 5CO2 (g) + 6H2O (l)
Delta H = -3505.8 kJ/mol

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Delta H = -393.5 kJ/mol

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Delta H = -286 kJ/mol

Possible answers:
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e. -178 kJ/mol
8 0
3 years ago
25 points! Help please
Sedbober [7]

Answer:

I think its b

Explanation:

but I wouldn't depend on this answer

4 0
3 years ago
Read 2 more answers
The enthalpy of formation of xef2(g) is –108 kj mol–1 and the bond dissociation enthalpy of the f–f bond is 155 kj mol–1 . what
S_A_V [24]
Xe +f2 →Xef2
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-108 k.s/mol =155 k.s/mol - 2B.Dxe-f
263kJ/mol/2 = B. D xe-f
B.D xef = 131.5 kJ/mol
132 kJ/mol
4 0
3 years ago
Read 2 more answers
Time that is measured in definite periods such as minutes, day,and years is called
aleksley [76]

Answer:

1. relative time

2.absolute time

Explanation:

Time that is measured in definite periods such as minutes, days, and years is called

a. Relative time.

b. absolute time.

7 0
3 years ago
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