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svp [43]
3 years ago
8

In an chemicalbreaction involving Fe and S, it was found that 45.2 g of Fes was produced. If the percent yield of the reaction i

s 94.5%, what is the theoretical yield of this reaction?​
Chemistry
1 answer:
saw5 [17]3 years ago
7 0

Answer:

47.8 g

Explanation:

Remember the equation for percent yield:

% yield = actual / theoretical

We're given two of the values in the question, so plug n' play:

0.945 = 45.2 / theoretical

theoretical = 47.8 g

Keep in mind you can use mass here without converting to moles because we're working with products only. If you were given a mass of reactants, you would need to convert to moles and using a balanced chemical equation find the corresponding moles of product produced.

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The chemical formula for ferric sulfate is Fe(SO4)3. Determine the following:
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Answer :

(a) The number of sulfur atoms are, 31.61\times 10^{23}.

(b) The mass of the mass of Fe_2(SO_4)_3 is, 1059.682 grams.

(c) The number of moles of Fe_2(SO_4)_3 is, 8.63\times 10^{-3}mole

(d) The mass of the mass of Fe_2(SO_4)_3 is, 19.95\times 10^{-22}g

Explanation :

(a) As we are given the number of moles of Fe_2(SO_4)_3 is, 1.75 mole. Now we have to calculate the number of sulfur atoms.

In the Fe_2(SO_4)_3, there are 2 iron atoms, 3 sulfur atoms, 12 oxygen atoms.

As, 1 mole of Fe_2(SO_4)_3 contains 3\times 6.022\times 10^{23} number of sulfur atoms.

So, 1.75 mole of Fe_2(SO_4)_3 contains 1.75\times 3\times 6.022\times 10^{23}=31.61\times 10^{23} number of sulfur atoms.

The number of sulfur atoms are, 31.61\times 10^{23}

(b) As we are given the number of moles of Fe_2(SO_4)_3 is, 2.65 mole. Now we have to calculate the mass of Fe_2(SO_4)_3.

\text{Mass of }Fe_2(SO_4)_3=\text{Moles of }Fe_2(SO_4)_3\times \text{Molar mass of }Fe_2(SO_4)_3

The molar mass of Fe_2(SO_4)_3 = 399.88 g/mole

\text{Mass of }Fe_2(SO_4)_3=2.65mole\times 399.88g/mole=1059.682g

The mass of the mass of Fe_2(SO_4)_3 is, 1059.682 grams.

(c) As we are given the mass of Fe_2(SO_4)_3 is, 3.45 grams. Now we have to calculate the moles of Fe_2(SO_4)_3.

\text{Mass of }Fe_2(SO_4)_3=\text{Moles of }Fe_2(SO_4)_3\times \text{Molar mass of }Fe_2(SO_4)_3

The molar mass of Fe_2(SO_4)_3 = 399.88 g/mole

3.45g=\text{Moles of }Fe_2(SO_4)_3\times 399.88g/mole

\text{Moles of }Fe_2(SO_4)_3=8.63\times 10^{-3}mole

The number of moles of Fe_2(SO_4)_3 is, 8.63\times 10^{-3}mole

(d) As we are given the formula unit of Fe_2(SO_4)_3 is, 3. Now we have to calculate the mass of Fe_2(SO_4)_3.

As we know that 1 mole of Fe_2(SO_4)_3 contains 6.022\times 10^{23} formula unit.

Formula used :

\text{Formula unit of }Fe_2(SO_4)_3=\text{Moles of }Fe_2(SO_4)_3\times 6.022\times 10^{23}

3=\text{Moles of }Fe_2(SO_4)_3\times 6.022\times 10^{23}

\text{Moles of }Fe_2(SO_4)_3=4.989\times 10^{-24}mole

Now we have to calculate the mass of Fe_2(SO_4)_3.

\text{Mass of }Fe_2(SO_4)_3=\text{Moles of }Fe_2(SO_4)_3\times \text{Molar mass of }Fe_2(SO_4)_3

The molar mass of Fe_2(SO_4)_3 = 399.88 g/mole

\text{Mass of }Fe_2(SO_4)_3=4.989\times 10^{-24}mole\times 399.88g/mole=19.95\times 10^{-22}g

The mass of the mass of Fe_2(SO_4)_3 is, 19.95\times 10^{-22}g

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