Question

I don"t any wrong answer you wasted my 300 point so gave this 1)1.84 gram mixture of CaCO3 and MgCO3 isl heated gave 0.96g. What is the % Mgco3 composition of mixture also gave chemical equation

Answer 1

Answer:

Please find below the solution to your problem.

Molar mass of CaCO3 = 100g/mol

Molar mass of MgCO3 = 84.3 g/mol

Let number of moles of CaCO3 and MgCO3 in the mixture is X & Y respectively .

So

100X + 84.3Y= 1.84.......................(1)

On heating the mixture...

CO2 goes off from the product but mixture of CaO and MgO remains in the solid form.

Molar mass of CaO = 56g/mole

Molar mass of MgO = 40.3 g/mole

From the equation it is clear that number of moles of MgO & CaO in product mixture is same as that of respective carbonates in the reactant mixture.

so

56X + 40.3Y = 0.96........................(2)

Solving equation 1 and 2 we get.

Y = 0.01

X = 0.009

Thus mass of CaCO3 in mixture = 100 X 0.0098 =0.98 g

Mass of MgCO3 in mixture = 84.3 X 0.01= 0.86 g

Mass % of CaCO3 in mixture=

= 0.98 / (0.98 + 0.86) × 100 = 53.26%

Mass % of MgCO3 in mixture =

100-53.26= 46.74%

Answer 2

$\bf{\huge{\underline{Question}}}$

1.84g mixture of CaCO3 and MgCO3 was heated to a constant weight to give 0.96g residue. What is the % Mgco3 composition of mixture also gave chemical equation.

$\bf{\huge{\underline{Answer}}}$

Let the mass of $CaCO_{3}$ be $x$g

So,the mass of $MgCO_{3}$ will be $(1.84 - x)$g

Now, $x$ g of $CaCO_{3}$ will give Cao $= \frac{56x}{100}$g

And, $(1.84 - x)$g of $MgCO_{3}$ will give MgO $= \frac{40(1.84-x)}{84}$g

Total mass of residue is 0.96g

So,$\frac{0.56+40(1.84-X)}{84}$ g= 0.96 g.

$x$=1g

Hence, % composition of $CaCO_{3}$ $= \frac{1}{1.84} \times 100 \\ =54.35\%$

And,% composition of $MgCO_{3}$ $=100\%-54.35\% \\ =45.65\%$