Question

20 g of magnesium carbonate sample decomposes on heating to giv carbon dioxide and 8 g of magnesium oxide . what will be the percentage purity in this sample//?

Answer 1

Thanks for asking the question!

ANSWER::

MgCO₃ ⇒ MgO + CO₂
84g          40g
y g             8 g (y is variable used)

y=(84 x 8)/40 = 16.8 g

Percentage purity of MgCO₃  = (16.8 x 100)/20=84%

Hope it helps!

Answer 2

percent purity-

according to the question -

MgCO3 ---------> MgO + CO2

now mass of MgCO3 - 24+12+16*3

= 48 g

similarly mass of CO2 - 44 g and

mass of MgO - 40 g

>>>>>in the following reaction 1 mole of magnesium carbonate forms one mole of magnesium oxide

and mole of MgO - 8/40

(mass/molecular mass)

thus 8/40 mole of magnesium carbonate will produced

1 mole of MgO - - - - - - > 1 mole of MgCO3

therefore -

8/40 = 1/5 mole - - - - - >1/5 mole of MgCO3

Now mass = 1/5 * (mass of MgCO3)

that is =1/5*84

=16.8

And percent purity = pure mass / impure mass *100

= 16.8 / 20 *100

=84%