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GalinKa [24]
3 years ago
10

If 40.0 g of HCl react with an excess of magnesium metal, what is the theoretical yield of hydrogen?

Chemistry
2 answers:
Bezzdna [24]3 years ago
6 0

Answer is A, 1.11 g .

mamaluj [8]3 years ago
5 0
Balanced equation: Mg+2HCl=MgCl2 + H2
Showing that if Mg is not a limiting factor then 2 moles of HCl on complete reaction liberate 1 mole of Hydrogen
therefore:
1 mole of HCl=35.5g, 40g=x
X= 40/35.5= 1.127mole
2 moles of HCl = 1 moles of Hydrogen
1.127=X
X=1.127/2 = 0.56
the theoretical yield of hydrogen
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Next, the balanced equation for the reaction. This is given below:

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From the balanced equation above,

1 mole of N2 reacted with 3 moles of H2 to produce 2 moles of NH3.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

1 mole of N2 reacted with 3 moles of H2.

Therefore, 0.83 moles will react with = (0.83 x 3) = 2.49 moles of H2.

From the calculations made above, we can see that only 2.49 moles out of 11.6 moles of H2 is required to react completely with 0.83 mole of N2.

Therefore, N2 is the limiting reactant.

Finally, we shall determine the maximum amount of NH3 produced from the reaction.

In this case, we shall use the limiting reactant because it will give the maximum yield of NH3 since all of it is consumed in the reaction.

The limiting reactant is N2 and the maximum amount of NH3 produced can be obtained as follow:

From the balanced equation above,

1 mole of N2 reacted to produce 2 moles of NH3.

Therefore, 0.83 mole of N2 will react to produce = (0.83 x 2) = 1.66 moles of NH3.

Therefore, the maximum amount of NH3 produced from the reaction is 1.66 moles.

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3 years ago
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