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

For the answer to the question above, let us assume that all Co2 is given off and this are the mass that is lost:

mass at start = 14.00 = 56.70 = 70.70g 
mass at end = 64.96g 
mass lost = 5.74g 

I hope my answer helped you. Have a nice day!

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 $HCl$.

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

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

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 $0.14\times 2=0.28$ 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 $CO_2$.

$\text{Mass of }CO_2=\text{Moles of }CO_2\times \text{Molar mass of }CO_2$

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

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