All categories

3 years ago

What is the mass of CO2 lost at 20 min from the limestone sample?

Chemistry

1 answer:

harkovskaia [24]3 years ago

8 0

Answer:

The mass, CO2 and CO3 from the limestone sample is discussed below in details.

Explanation:

(A) mass loss of sample of limestone after 20 min

= 0.8437g-0.5979g = 0.2458 g

From the given reaction of limestone, 2 mol of the sample gives 2 moles of CO 2.

Therefore

184.4 g ( molar mass of limestone) gives2× 44 g of carbon dioxide.

1 g of sample gives 88/184.4 g of carbon dioxide

Hence 0.2458 g sample gives

= 88/184.4 × 0.2458 g = 0.117 g carbon dioxide

(B) mole of CO 2 lost = weight/ molar mass

= 0.117 g / 44 g/mol =0.0027 mole

(C). 1 mol of limestone contain 2 mol of carbonate ion

From the reaction we know that carbonate ion of limestone is converted into carbondioxide

Hence lost carbonate ion = 0.2458 g

(D) we know that

1 mol limestone contain 1mol CaCO 3

Hence in sample present CaCO 3

= 1mole / 184.4 g × 0.8437 g= 0.00458 mol CaCO3

You might be interested in

Various members of a class of compounds, alkenes, react with hydrogen to produce a corresponding alkane. Termed hydrogenation, t

Vitek1552 [10]

<u>Answer:</u> The mass of decane produced is $1.743\times 10^2g$

<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)

Mass of hydrogen gas = 2.45 g

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 1:, we get:

$\text{Moles of }H_2=\frac{2.45g}{2g/mol}=1.225mol$

The chemical equation for the hydrogenation of decene follows:

$C_{10}H_{20}(l)+H_2(g)\rightarrow C_{10}H_{22}(s)$

As, decene is present in excess. So, it is considered as an excess reagent.

Thus, hydrogen gas is a limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

1 mole of hydrogen gas produces 1 mole of decane.

So, 1.225 moles of hydrogen gas will produce = $\frac{1}{1}\times 1.225=1.225mol$ of decane

Now, calculating the mass of decane by using equation 1, we get:

Moles of decane = 1.225 mol

Molar mass of decane = 142.30 g/mol

Putting values in equation 1, we get:

$1.225mol=\frac{\text{Mass of decane}}{142.30g/mol}\\\\\text{Mass of carbon dioxide}=(1.225mol\times 142.30g/mol)=174.3g=1.743\times 10^2g$

Hence, the mass of decane produced is $1.743\times 10^2g$

5 0

3 years ago

Hydrogen is a reactive gas under normal conditions.Based on this property alone, to which group of the periodic table should hyd

denis23 [38]

Even though Hydrogen is originally in group 1, based on this property, we can say it is in group 6.

Because:

Group 6 would mean that it only needs 2 more valence electrons till the octet (8 valence electrons). This would make it reactive, yet, in normal conditions, unlike group 7.

5 0

4 years ago

Read 2 more answers

Sometimes clothes removed from the dryer cling together. What kinds of charges are on the clothes ?

inn [45]

The answer is "opposite charges."

4 0

3 years ago

A diver has a lung capacity of 2.4 L when the pressure is 0.8 atm. What is the volume of the diver’s lungs when the pressure cha

tangare [24]

Answer:

The final volume is 1.6 L.

Explanation:

It is given that,

A diver has a lung capacity of 2.4 L when the pressure is 0.8 atm. We need to find the volume of the diver’s lungs when the pressure changes to 1.2 atm. Let V₂ is volume.

It is based on Boyle's law. According to this law,

$PV=K$