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When 3.1g of a metal carbonate XC03 were heated to constant mass,2.0g of oxide of meta were formed. on reduction of this oxide ,1.6g of...

When 3.1g of a metal carbonate XC03 were heated to constant mass,2.0g of oxide of meta were formed. on reduction of this oxide ,1.6g of metal remained .calculate the atomic number of X?

Answers

Masinde

XC03=XO+CO2
mass of XCO3=3.1
mass of XO =2.0
mass of CO2 =1.1

moles of C02 produced=1.1/44
=0.025moles
Sinse the mole ratio of the reaction is 1:1,moles of XO produced is 0.025.
XO will be reduced as,
X0 + H2 = X + H20
mole ratio of reaction is 1:1
therefore, the moles of X produced is 0.025
RAM of X= mass/ moles
=1.6/0.025
=64
Therefore the atomic number is 32.
Mr Masinde answered the question on May 5, 2018 at 08:37

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