Question

You carefully weigh out 14.00 g of CaCO3 powder and add it to 56.70 g of HCl solution. You notice bubbles as a reaction takes place. You then weigh the resulting solution and find that it has a mass of 64.96 g . The relevant equation is

CaCO3(s)+2HCl(aq)→H2O(l)+CO2(g)+CaCl2(aq)
Assuming no other reactions take place, what mass of CO2 was produced in this reaction?

Answer 1

The mass of CO2 produced in the reaction is 5.74 g.

Step-by-step explanation:

To calculate the mass of CO2 produced in the reaction, we need to determine the change in mass caused by the formation of CO2. We start with the mass of CaCO3 (14.00 g) and the mass of HCl solution (56.70 g). The total mass of the reactants is the sum of these two: 14.00 g + 56.70 g = 70.70 g.

The mass of the resulting solution is 64.96 g, so the change in mass is 70.70 g - 64.96 g = 5.74 g. This change in mass corresponds to the mass of CO2 produced in the reaction, so the mass of CO2 is 5.74 g.

Answer 2

Answer : The mass of
`CO_2` produced will be, 6.16 grams.

Explanation : Given,

Mass of
`CaCO_3` = 14 g

Mass of
`HCl` = 56.70 g

Molar mass of
`CaCO_3` = 100 g/mole

Molar mass of
`HCl` = 36.5 g/mole

Molar mass of
`CO_2` = 44 g/mole

First we have to calculate the moles of
`CaCO_3` and
`HCl`.

`\text{Moles of }CaCO_3=\frac{\text{Mass of }CaCO_3}{\text{Molar mass of }CaCO_3}=(14g)/(100g/mole)=0.14moles`

`\text{Moles of }HCl=\frac{\text{Mass of }HCl}{\text{Molar mass of }HCl}=(56.70g)/(36.5g/mole)=1.55moles`

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

`CaCO_3(s)+2HCl(aq)\rightarrow CO_2(g)+H_2O(l)+CaCl_2(aq)`

From the balanced reaction we conclude that

As, 1 moles of
`CaCO_3` react with 2 mole of
`HCl`

So, 0.14 moles of
`CaCO_3` react with
`0.14* 2=0.28` moles of
`HCl`

From this we conclude that,
`HCl` is an excess reagent because the given moles are greater than the required moles and
`CaCO_3` is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of
`CO_2`.

As, 1 moles of
`CaCO_3` react to give 1 moles of
`CO_2`

So, 0.14 moles of
`CaCO_3` react to give 0.14 moles of
`CO_2`

Now we have to calculate the mass of
`CO_2`.

`\text{Mass of }CO_2=\text{Moles of }CO_2* \text{Molar mass of }CO_2`

`\text{Mass of }CO_2=(0.14mole)* (44g/mole)=6.16g`

Therefore, the mass of
`CO_2` produced will be, 6.16 grams.