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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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Shkiper50 [21]

Answer: Theoretical mass of sodium sulphate ( Na_2SO_4 ) is 514.118 grams.

Explanation: For a given reaction,

2NaOH(aq.)+H_2SO_4(aq.)\rightarrow Na_2SO_4(s)+2H_2O(l)

As NaOH is used in excess, therefore it is an excess reagent and H_2SO_4 is a limiting reagent as the quantity of the product will depend on it.

We are given 355 grams of H_2SO_4.

Molar mass of H_2SO_4 = 98.079 g/mol

Molar mass of Na_2SO_4 = 142.04 g/mol

1 mole of H_2SO_4 is producing 1 mole of Na_2SO_4, so

98.079 g/mol of  H_2SO_4 will produce 142.04 g/mol of Na_2SO_4

355 grams of H_2SO_4 will produce = 142.04g/mol \times \frac{355g}{98.079g/mol} of Na_2SO_4

Mass of Na_2SO_4 = 514.118 grams


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Explanation:

<em>A chemist dissolves 716. mg of pure potassium hydroxide in enough water to make up 130. mL of solution. Calculate the pH of the solution. (The temperature of the solution is 25 °C.) Be sure your answer has the correct number of significant digits.</em>

Step 1: Given data

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Step 3: Calculate the molar concentration of KOH

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The molar ratio of KOH to OH⁻is 1:1. Then, [OH⁻] = 0.0985 M

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We will use the following expression.

pOH = -log [OH⁻] = -log 0.0985 = 1.01

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