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Margaret [11]
3 years ago
8

How many grams of ZnCl2 will be produced from 26.0g of Zn and 42.0g of HCl. Round to three significant figures

Chemistry
1 answer:
bija089 [108]3 years ago
3 0

<u>Answer:</u> The mass of zinc chloride that can be produced is 54.1 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)

  • <u>For zinc:</u>

Given mass of zinc = 26.0 g

Molar mass of zinc = 65.4 g/mol

Putting values in equation 1, we get:

\text{Moles of zinc}=\frac{26g}{65.4g/mol}=0.397mol

  • <u>For HCl gas:</u>

Given mass of HCl gas = 42 g

Molar mass of HCl gas = 36.5 g/mol

Putting values in equation 1, we get:

\text{Moles of HCl}=\frac{42g}{36.5g/mol}=1.15mol

The chemical equation for the reaction of zinc and chlorine gas follows:

Zn+2HCl\rightarrow ZnCl_2+H_2

By Stoichiometry of the reaction:

1 mole of zinc reacts with 2 moles of HCl

So, 0.397 moles of zinc will react with = \frac{2}{1}\times 0.397=0.794mol of HCl

As, given amount of HCl is more than the required amount. So, it is considered as an excess reagent.

Thus, zinc is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of zinc produces 1 mole of zinc chloride

So, 0.397 moles of zinc will produce = \frac{1}{1}\times 0.397=0.397moles of zinc chloride

Now, calculating the mass of zinc chloride from equation 1, we get:

Molar mass of zinc chloride = 136.3 g/mol

Moles of zinc chloride = 0.397 moles

Putting values in equation 1, we get:

0.397mol=\frac{\text{Mass of zinc chloride}}{136.3g/mol}\\\\\text{Mass of zinc chloride}=(0.397mol\times 136.3g/mol)=54.1g

Hence, the mass of zinc chloride that can be produced is 54.1 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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