# A 1.871 gram sample of an unknown metallic carbonate is decomposed by heating to form the metallic oxide and 0.656 g of carbon dioxide according to the equation MCO3(s) -> MO(s) + CO2(g) What...

A 1.871 gram sample of an unknown metallic carbonate is decomposed by heating to form the metallic oxide and 0.656 g of carbon dioxide according to the equation

MCO3(s) -> MO(s) + CO2(g)

What is the metal?

jeew-m | College Teacher | (Level 1) Educator Emeritus

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Mole mass in `g/(mol)`

`C=12`

`O=16`

`MCO_3rarr MO+CO_2`

Mole ratio

`CO_2 : MCO_3 =1:1`

Amount of `CO_­2` produced `=0.656/(12+32) =0.015`

Amount of `MCO_3`reacted =0.015 mol

It is given that 1.871g of the unknown metal carbonate reacted. If the molar mass of M is X g/mol.

`(X+12+16xx3)xx0.015 =1.871`

`X=64.73`

According to the molar mass the metal the metal can be Copper (Cu=63.5) Or Zinc(Zn=65.4).

Assumptions

• The metal carbonate sample is pure and without impurities .
• The sample is fully decomposed at the heating process.
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