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SashulF [63]
3 years ago
5

0.100 mol of CaCO3 and 0.100 mol CaO are placed in an 10.0 L evacuated container and heated to 385 K. When equilibrium is reache

d the pressure of CO2 is 0.220 atm. 0.300 atm of CO2 is added, while keeping the temperature constant and the system is allowed to reach again equilibrium. What will be the final mass of CaCO3
Chemistry
1 answer:
HACTEHA [7]3 years ago
4 0

Answer:

12.531 grams will be the final mass of calcium carbonate.

Explanation:

CaCO_3(s)\rightleftharpoons CaO(s)+CO_2(g)

Partial pressure of carbon dioxide gas at equilibrium =P=0.220 atm

Volume of carbon dioxide gas = V = 10.0 L

Moles of carbon dioxide gas formed = n

Temperature of the gas = T = 385 K

PV=nRT( ideal gas equation)

n=\frac{PV}{RT}=\frac{0.220 atm\times 10.0 L}{0.0821 atm L/mol K\times 385 K}=0.06960 mol

According to reaction , 1 mole of carbon dioxide gas is formed from 1 mole of calcium carbonate,0.06960 mole of carbon dioxide gas will be obtained from :

\frac{1}{1}\times 0.06960 mol=0.06960 mol calcium carbonate

Moles of calcium carbonate at equilibrium = 0.100 mol - 0.06960 mol = 0.03040 mol

After addition of 0.300 atm of carbon dioxide gas, more amount of calcium carbonate will be be formed.

Here, at the same temperature, the equilibrium pressure of the carbon dioxide gas is 0.220 atm so, entire 0.300 atm of carbon dioxide will get convert to calcium carbonate.

So amount moles of carbon dioxide gas added = moles of calcium carbonate formed after re-establishment of an equilibrium :

Partial pressure of carbon dioxide gas added at equilibrium =P=0.300 atm

Volume of carbon dioxide gas = V = 10.0 L

Moles of carbon dioxide gas formed = n

Temperature of the gas = T = 385 K

PV=nRT( ideal gas equation)

n=\frac{PV}{RT}=\frac{0.300 atm\times 10.0 L}{0.0821 atm L/mol K\times 385 K}=0.09491 mol

Moles of calcium carbonate = 0.09491 mol

Total moles of calcium carbonate in the container :

=0.03040 mol + 0.09491 mol = 0.12531 mol

Mass of 0.12531 moles of calcium carbonate :

0.12531 mol × 100 g/mol = 12.531 g

12.531 grams will be the final mass of calcium carbonate.

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Question List (4 items) (Drag and drop into the appropriate area) Find the volume of HCl that will neutralize the base. Find the
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The question is incomplete, the complete question is:

The solubility of slaked lime, Ca(OH)_2, in water is 0.185 g/100 ml. You will need to calculate the volume of 2.50\times 10^{-3}M HCl needed to neutralize 14.5 mL of a saturated

<u>Answer:</u> The volume of HCl required is 290mL, the mass of Ca(OH)_2 is 0.0268g, the moles of

<u>Explanation:</u>

Given values:

Solubility of Ca(OH)_2 = 0.185 g/100 mL

Volume of Ca(OH)_2 = 14.5 mL

Using unitary method:

In 100 mL, the mass of Ca(OH)_2 present is 0.185 g

So, in 14.5mL. the mass of Ca(OH)_2 present will be =\frac{0.185}{100}\times 14.5=0.0268g

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

The equation used is:

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

Given mass of Ca(OH)_2 = 0.0268 g

Molar mass of Ca(OH)_2 = 74 g/mol

Plugging values in equation 1:

\text{Moles of }Ca(OH)_2=\frac{0.0268g}{74g/mol}=0.000362 mol

Moles of OH^- present = (2\times 0.000362)=0.000724mol

The chemical equation for the neutralization of calcium hydroxide and HCl follows:

Ca(OH)_2+2HCl\rightarrow CaCl_2+2H_2O

By the stoichiometry of the reaction:

Moles of OH^- = Moles of H^+ = 0.000724 mol

The formula used to calculate molarity:

\text{Molarity of solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (mL)}} .....(2)

Moles of HCl = 0.000724 mol

Molarity of HCl = 2.50\times 10^{-3}

Putting values in equation 2, we get:

2.50\times 10^{-3}mol=\frac{0.000724\times 1000}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{0.000725\times 1000}{2.50\times 10^{-3}}=290mL

Hence, the volume of HCl required is 290mL, the mass of Ca(OH)_2 is 0.0268g, the moles of

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Why does water boils at a higher temperature than a non-polar solvent like ether?
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<span>a large amount of energy is released when water dissociates into oppositely charged ions.</span>
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<span>You can see that there are 2 moles of gas in the reactants and 3 moles of gas in the products. </span>

<span>1 moles of ideal gas occupies the same volume as 1 mole of any other ideal gas under the same conditions of temp and pressure. </span>

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