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Mademuasel [1]
3 years ago
12

How many grams of h2 gas can be produced by the reaction of 63.0 grams of al(s) with an excess of dilute hydrochloric acid in th

e reaction shown below? 2 al(s) + 6 hcl(aq) → 2 alcl3(aq) + 3 h2(g)?
Chemistry
2 answers:
vitfil [10]3 years ago
7 0
How many grams of h2 gas can be produced by the reaction of 63.0 grams of al(s) with an excess of dilute hydrochloric acid in the reaction shown below? 2 al(s) + 6 hcl(aq) → 2 alcl3(aq) + 3 h2(g)?
Sergeeva-Olga [200]3 years ago
5 0

<u>Answer:</u> The mass of hydrogen gas produced by the reaction is 6.9 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}      .....(1)

Given mass of aluminium = 63 g

Molar mass of aluminium = 27 g/mol

Putting values in equation 1, we get:

\text{Moles of aluminium}=\frac{63g}{27g/mol}=2.33mol

For the given chemical reaction:

2Al(s)+6HCl(aq.)\rightarrow 2AlCl_3(aq.)+3H_2(g)

Hydrochloric acid is present in excess. So, it is considered as an excess reagent. And, aluminium metal is a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of aluminium metal produces 3 moles of hydrogen gas.

So, 2.33 moles of aluminium metal will produce = \frac{3}{2}\times 2.33=3.45mol of hydrogen gas

Now, calculating the mass of hydrogen gas by using equation 1:

Moles of hydrogen gas = 3.45 moles

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 1, we get:

3.45mol=\frac{\text{Mass of hydrogen gas}}{2g/mol}\\\\\text{Mass of hydrogen gas}=(3.45mol\times 2g/mol)=6.9g

Hence, the mass of hydrogen gas produced by the reaction is 6.9 grams

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Ethyl butyrate, CH3CH2CH2CO2CH2CH3, is an artificial fruit flavor commonly used in the food industry for such flavors as orange
SIZIF [17.4K]

Answer:

A. 10.0 grams of ethyl butyrate would be synthesized.

B. 57.5% was the percent yield.

C. 7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

Explanation:

CH_3CH_2CH_2CO_2H(l)+CH_2CH_3OH(l)+H^+\rightarrow CH_3CH_2CH_2CO_2CH_2CH_3(l)+H_2O(l)

A

Moles of butanoic acid = \frac{7.60 g}{88 g/mol}=0.08636 mol

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

\frac{1}{1}\times 0.08636 mol=0.08636 mol of ethyl butyrate

Mass of 0.08636 moles of ethyl butyrate =

0.08636 mol × 116 g/mol = 10.0 g

Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 100%

Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

100\%=\frac{\text{Experimental yield}}{10.0 g}\times 100

Experimental yield = 10.0 g

10.0 grams of ethyl butyrate would be synthesized.

B

Theoretical yield of ethyl butyrate  = 10.0 g

Experimental yield ethyl butyrate = 5.75 g

Percentage yield of the reaction = ?

Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

=\frac{5.75 g}{10.0 g}\times 100=57.5\%

57.5% was the percent yield.

C

Moles of butanoic acid = \frac{7.60 g}{88 g/mol}=0.08636 mol

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

\frac{1}{1}\times 0.08636 mol=0.08636 mol of ethyl butyrate

Mass of 0.08636 moles of ethyl butyrate =

0.08636 mol × 116 g/mol = 10.0 g

Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 78.0%

Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

78.0\%=\frac{\text{Experimental yield}}{10.0 g}\times 100

Experimental yield = 7.80 g

7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

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