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Darya [45]
4 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]4 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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3 years ago
For the following reaction, 9.60 grams of butane (C4H10) are allowed to react with 17.0 grams of oxygen gas. butane (C4H10) (g)
dangina [55]

Answer:

14.4 g of CO_{2} can be produced.

Explanation:

Balanced equation: 2C_{4}H_{10}+13O_{2}\rightarrow 8CO_{2}+10H_{2}O

                                        Molar mass (g/mol)

                  C_{4}H_{10}                    58.12

                    O_{2}                         32

                  CO_{2}                       44.01

So, 9.60 g of C_{4}H_{10} = \frac{9.60}{58.12}mol=0.165mol

      17.0 g of O_{2} = \frac{17.0}{32}mol=0.531mol

According to balanced equation-

2 moles of C_{4}H_{10} produce 8 moles of CO_{2}

So, 0.165 moles of C_{4}H_{10} produce (\frac{8}{2}\times 0.165)mol of CO_{2}  or 0.660 moles of CO_{2}

13 moles of O_{2} produce 8 moles of CO_{2}

So, 0.531 moles of O_{2} produce (\frac{8}{13}\times 0.531)moles of CO_{2} or 0.327 moles of CO_{2}

As least number of moles of CO_{2} are produced from O_{2} therefore O_{2} is the limiting reagent.

So, maximum amount of CO_{2} that can be formed = 0.327 moles

                                                                              = (0.327\times 44.01)g

                                                                              = 14.4 g

3 0
3 years ago
which term is used when sugar is completely dissolved in water                                                                 
GaryK [48]
The term which is used is homogeneous.
when sugar is completely dissolved in the water, the mixture or solution homogeneous, both in same phase and same uniform texture that is liquid.
There two types of mixtures are heterogeneous and homogeneous in different phases.
If sugar is not completely dissolved in water and you see the crystals of sugar in water, then the solution will be heterogeneous.
3 0
4 years ago
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