Question

How many grams of carbon-di-oxide gas will be produced when 16.8 grams of 90% magnesium carbonate reacts with excess hydrochloric acid? [ Mg -24u, C-12u, O-16u]​

Answer 1

Answer:

We can calculate the percentage composition by dividing the mass of a component by the total mass of the mixture. Then the ratio is to be multiplied by 100. It is also called the mass percent.

Complete step by step solution:

- We will write the reaction first:

CaCO3+2HCl→CaCl2+H2O

Here, we will consider the moles of calcium carbonate as X mole and hydrochloric acid as 2x.

MgCO3+2HCl→MgCl2+H2O

Here, we will consider moles of magnesium carbonate as Y mole and hydrochloric acid as 2y.

Now, we can write

N1V1−N2V21000

where, N1

= Number of moles of HCl

V1

= Volume of HCl

N2

= Number of moles of NaOH

V2

= Volume of NaOH

By putting all values in above equation, we get:

2x+2y=(0.8×50−0.25×16)10002x+2y=(40−4)10002x+2y=3610002x+2y=0.036x+y=0.0362x+y=0.018

Consider this equation as equation 1,

Now, we will find the molecular weight of calcium carbonate:

CaCO3=40+12+3×16=40+12+48=40+60=100

Now, we will find the molecular weight of magnesium carbonate:

MgCO3=24+12+3×16=24+12+48=24+60=84

We can write:

x×100+y×84=1.64

Consider this equation as equation 2

- From equation 1 and 2 we can find the value of x as:

from equation 1 we can write: x + y =0.018

so, y = x + 0.018

Now substitute the value of y in equation 2:

100(x)+84(0.018−x)=1.64100x+1.512−84x=1.64100x−84x+1.512=1.6416x=1.64−1.51216x=0.128x=0.12816x=0.008

Now, we will write the percentage of calcium carbonate as:

Weight of calcium carbonate x =0.008 gm

%ofCaCO=x×molecular weight1.64×100

=0.008×1001.64×100

=0.81.64×100

=0.4878×100

=48.78

Now, we will write the percentage of magnesium carbonate as:

100 - percentage of calcium carbonate

= 100- 48.78

=51.22%