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Nesterboy [21]
3 years ago
15

What is the concentration of chloride in a solution made with 0.808 grams of CaCl2 and 250.0 ml of water.?

Chemistry
1 answer:
densk [106]3 years ago
8 0

Answer:

0.0584 M

Explanation:

From the question given above, the following data were obtained:

Mass of CaCl₂ = 0.808 g

Volume of water = 250 mL

Concentration of chloride =?

Next, we shall determine the number of mole in 0.808 g of CaCl₂. This can be obtained as follow:

Mass of CaCl₂ = 0.808 g

Molar mass of CaCl₂ = 40 + (35.5 × 2)

= 40 + 71

= 111 g/mol

Mole of CaCl₂ =?

Mole = mass / Molar mass

Mole of CaCl₂ = 0.808 / 111

Mole of CaCl₂ = 0.0073 mole

Next, we shall convert 250 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

250 mL = 250 mL × 1 L / 1000 mL

250 mL = 0.25 L

Next, we shall determine the molarity of CaCl₂. This can be obtained as follow:

Mole of CaCl₂ = 0.0073 mole

Volume of water = 0.25 L

Molarity of CaCl₂ =?

Molarity = mole /Volume

Molarity of CaCl₂ = 0.0073 / 0.25

Molarity of CaCl₂ = 0.0292 M

Finally, we shall determine the concentration of the chloride as illustrated below:

CaCl₂ <=> Ca²⁺ + 2Cl¯

From the equation above,

1 mole of CaCl₂ produced 2 mole of Cl¯.

Therefore, 0.0292 M CaCl₂ will produce = 0.0292 × 2 = 0.0584 M Cl¯.

Thus, the concentration of the chloride ion (Cl¯) in the solution is 0.0584 M

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<u>Answer:</u> The percent yield of the reaction is 91.8 %

<u>Explanation:</u>

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

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Putting values in equation 1, we get:

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  • <u>For oxygen gas:</u>

Given mass of oxygen gas = 10.0 g

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Putting values in equation 1, we get:

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The chemical equation for the reaction of B_5H_9 and oxygen gas follows:

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By Stoichiometry of the reaction:

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So, 0.3125 moles of oxygen gas will react with = \frac{2}{12}\times 0.3125=0.052mol of B_2H_5

As, given amount of B_2H_5 is more than the required amount. So, it is considered as an excess reagent.

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

By Stoichiometry of the reaction:

12 moles of oxygen gas produces 5 moles of B_2O_3

So, 0.3125 moles of oxygen gas will produce = \frac{5}{12}\times 0.3125=0.130moles of water

Now, calculating the mass of B_2O_3 from equation 1, we get:

Molar mass of B_2O_3 = 69.93 g/mol

Moles of B_2O_3 = 0.130 moles

Putting values in equation 1, we get:

0.130mol=\frac{\text{Mass of }B_2O_3}{69.63g/mol}\\\\\text{Mass of }B_2O_3=(0.130mol\times 69.63g/mol)=9.052g

To calculate the percentage yield of B_2O_3, we use the equation:

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

Experimental yield of B_2O_3 = 8.32 g

Theoretical yield of B_2O_3 = 9.052 g

Putting values in above equation, we get:

\%\text{ yield of }B_2O_3=\frac{8.32g}{9.052g}\times 100\\\\\% \text{yield of }B_2O_3=91.8\%

Hence, the percent yield of the reaction is 91.8 %

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=  0.412mol  x  27  g/mol  =  11.124  grams
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