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2 years ago

What is the theoretical yield of Li3N in grams when 12.8 g of Li is heated with 34.9 g of N2?

Chemistry

1 answer:

Olenka [21]2 years ago

4 0

Answer:- 21.4 grams of $Li_3N$ are formed.

Solution:- The balanced equation is:

$6Li+N_2\rightarrow 2Li_3N$

From this equation, lithium and nitrogen reacts in 6:1 mol ratio. Limiting reactant gives the theoretical yield of the product. We will calculate the grams of the product for the given grams of both the reactants and see which one of them gives the limited amount of the product. This limited amount of the product will be the theoretical yield.

The molar mass of Li is 6.94 gram per mol and for $N_2$ It is 28.02 gram per mol. The molar mass of $Li_3N$ is 34.83 gram per mol. The calculations for the grams of the product for given grams of both the reactants are shown below:

$12.8gLi(\frac{1molLi}{6.94gLi})(\frac{2molLi_3N}{6molLi})(\frac{34.83gLi_3N}{1molLi_3N})$

= $21.4gLi_3N$

$34.9gN_2(\frac{1molN_2}{28.02gN_2})(\frac{2molLi_3N}{1molN_2})(\frac{34.83gLi_3N}{1molLi_3N})$

= $86.8gLi_3N$

From above calculations, Li gives least amount of the product. So, 21.4 g of $Li_3N$ are formed.

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3 years ago

When 5.00 g of Al2S3 and 2.50 g of H2O are reacted according to the following reaction: Al2S3(s) + 6 H2O(l) → 2 Al(OH)3(s) + 3 H

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$Y=58.15\%$

Explanation:

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For the given chemical reaction:

$Al_2S_3(s) + 6 H_2O(l) \rightarrow 2 Al(OH)_3(s) + 3 H_2S(g)$

We first must identify the limiting reactant by computing the reacting moles of Al2S3:

$n_{Al_2S_3}=5.00gAl_2S_3*\frac{1molAl_2S_3}{150.158 gAl_2S_3} =0.0333molAl_2S_3$

Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:

$n_{Al_2S_3}^{consumed}=2.50gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molAl_2S_3}{6molH_2O}=0.0231mol Al_2S_3$

Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:

$m_{Al(OH)_3}=0.0231molAl_2S_3*\frac{2molAl(OH)_3}{1molAl_2S_3}*\frac{78gAl(OH)_3}{1molAl(OH)_3} =3.61gAl(OH)_3$