Question

a sample of mgco3 is 50% pure. if on heating 4.0g of mgo and 4.4 g co2 are formed . calculate amount of sample taken mgco3 for heating.

plz gave answer with explain .​

Answer 1

When the sample is 100% pure ,

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

(84g) ( 40g) ( 44g)

When sample is 100% pure 84g of MgCO3 produces 40g of MgO and 44g CO2 ( ∵ 1 mol gives 1 mol)

_____________________________

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

(42g) ( 20g) (22g)

when sample is 50% pure so 1 mole of MgCO3 will have 42g (84/2)

42g of MgCO3 gives 20g of MgO and 22g og CO2 { ∵ 1/2 mol og MgCO3 gives 1/2 moles }

According to the question ,

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

(x) ( 4g) ( 4.4g)

using law of conservation of mass ,

x = 4 + 4.4 = 8.4g

Answer 2

Answer:

hey result ka kya hal hai.. btao plz

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