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2 years ago

Can someone please help me with this question also explain the answers I am so confused thank you.

Chemistry

1 answer:

Archy [21]2 years ago

4 0

The theoretical yield of H₂S is 13.5 g.

The percent yield is 75.5 %.

<h3>What is the theoretical yield of H₂S from the reaction?</h3>

The equation of the reaction is given below:

FeS + 2 HCl → FeCl₂+ H₂S

Moles of FeS reacting = mass/molar mass

Molar mass of FeS = 88 g/mol

Moles of FeS reacting = 35/88 = 0.398 moles

Moles of H₂S produced = 0.398 moles

Molar mass of H₂S = 34 g/mol

Mass of H₂S produced = 0.398 * 34 = 13.5 g

Theoretical yield of H₂S is 13.5 g.

Percent yield = actual yield/theoretical yield * 100%

Actual yield of H₂S = 10.2 g

Percent yield = 10.2/13.5 * 100%

Percent yield = 75.5 %

In conclusion, the actual yield is less than the theoretical yield.

Learn more about percent yield at: brainly.com/question/8638404

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If 16.0 grams of aluminum oxide were actually produced, what is the percent yield of the reaction below given that you start wit

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Answer: The percent of yield is 50.03%.

Explanation:

First, we need to balance the equation:

$4Al + 3O_{2}$ ⇒ $2Al_{2}O_{3}$

We need to remember that the chemical equations are written in moles, so we have to convert the amounts in grams to moles, using the molecular weight of every compound: Al2O3 (101.96 g/mol), Al (29.98 g/mol) and O2 (31.99 g/mol).

In consequence, the amounts in moles of every compound will be:

$16 gAl_{2}O_{3} * \frac{1 mol}{101.96 g} =0.157 mol Al_{2}O_{3}$

$10 g Al * \frac{1mol}{29.98 g}= 0.333 mol Al$

$19 g O_{2}*\frac{1 mol}{31.99 g} =0.594 mol O_{2}$

Now, we have to find out which is the limit reagent or in other words, which of the reagents will be consumed first, taking into account the stoichiometric ratio of the balanced equation:

$0.333 mol Al * \frac{2 mol Al_{2}O_{3}}{4 mol Al} =0.1665 mol Al_{2}O_{3}$

$0.594 mol O_{2} * \frac{2 mol Al_{2}O_{3}}{3 mol O_{2}} =0.396 mol Al_{2}O_{3}$

As you can see, the maximum amount (theoretically) of Al2O3 that can be produced is 0.1665 mol.

Finally, we have to use the yield formula to calculate the percent yield of the reaction:

$Percent of yield = \frac{actual yield}{theoretical yield} * 100 = \frac{0.157 mol Al_{2}O_{3}}{0.1665 mol Al_{2}O_{3}} * 100 = 94.25$

Therefore, the percent of yield is 50.03%.

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