A mineral consists of an equimolar mixture of the carbonates of two bivalent metals. One metal is present to the extent of 12.5% by mass.2.8 g of the mineral on heating loat 1.32 of `CO_(2)`. What is the % by mass of the other metal ?

Options

A. 87.5
B. 35.71
C. 65.11
D. 23.21

Answer

D

Explanation

Let two equimolar carbonates are `ACO_(3)` and `BCO_(3)`
mass of metal A = `2.8 xx 0.125` = 0.35 g
`ACO_(3)+BCO_(3) to AO+BO+2CO_(2)`
Total two moles `CO_(3)^(2-)` provide 2 moles of `CO_(2)` or 60 g `CO_(3)^(2-)` provide 44 g of `CO_(2)`
Total mass of `CO_(3)^(2-)` required to produce 1.32g of `CO_(2)=(60)/(44) xx 1.32=1.8` g
mass of metal (A + B) = 2.8 - 1.8 = 1 g
mass of B =1 - 0.35 = 0.65 g
`"% of B"=(0.65)/(2.80) xx 100=23.21`