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A mixture contains equi molar quantities of carbonates of two bivalent metals. One metal is present to the extent of 13.5 %by weight in the mixture and 2.50 gram of the mixture on heating leaves a residue of 1.18 gram Calculate the percentage by weight of the other metal.

A mixture contains equi molar quantities of carbonates of two bivalent metals. One metal is present to the extent of 13.5 %by weight in the mixture and 2.50 gram of the mixture on heating leaves a residue of 1.18 gram  Calculate the percentage by weight of the other metal. 

Grade:11

1 Answers

Arun
25750 Points
5 years ago
XCO3 --> x = moles of XCO3 

YCO3 --> y = moles of YCO3 

I know: 

x = y 

XCO3 --> XO + CO2 
YCO3 --> YO + CO2 

moles CO2 = x+y = 1.233/PM(CO2) = 1.233 / 44 = 2.80*10^-2 
then x = y = 2.80*10^-2 / 2 = 1.40*10^-2 

x*PM(X) = 2.58 * (13.2/100) = 0.33354g 
--> PM(X) = 0.33354 / 1.40*10^-2 = 23.94 

--> PM(XCO3) = 23.94 + 12 + 3*16 = 83.94 

--> x*PM(XCO3) = mass of XCO3 = 1.40*10^-2 * 83.94 = 1.175g 


1.175g / 2.58g = 0.455 --> 45.5% of XCO3 

100 - 45.5 = 54.5% of YCO3

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