1 gm of a mixture of an equal number of moles of L i 2 C O 3 and M 2 C O 3 required 44.44 ml of 0.5 M HCl for completion of the reaction: L i 2 C O 3 + 2 HCl → 2 LiCl + H 2 O + C O 2 M

Question

1 gm of a mixture of an equal number of moles of $L i_{2} C O_{3}$ and $M_{2} C O_{3}$ required 44.44 ml of 0.5 M HCl for completion of the reaction:

$L i_{2} C O_{3} + 2 HCl \to 2 LiCl + H_{2} O + C O_{2}$

$M_{2} C O_{3} + 2 HCl \to 2 MCl + H_{2} O + C O_{2}$

If the atomic mass of Li is 7, and the atomic mass of metal 'M' is X amu, then the nearest integer value of $(M - 16)$ is:

TestHub · Accepted Answer

Question Explanation: Determining the atomic mass of a metal in a carbonate mixture.

Concept: Neutralization and Molarity.

Solution: An equimolar mixture of $Li_{2} CO_{3}$ and $M_{2} CO_{3}$ reacts with HCl . The total moles of HCl used is twice the total moles of carbonates. We find the moles of each carbonate from the titration data and then calculate the unknown atomic mass X .

Let each species be ' a ' moles & M be atomic mass of metal. $a \times (2 \times 7 + 12 + 48) + a (2 M + 12 + 48) = 1 \dots (i)$

And a moles of each carbonate reacts with 2 a moles of HCl .

Hence $\to 4 a = (44.44 \times 0.5 \times 1 0^{- 3})$

$a = 11.11 \times 0.5 \times 1 0^{- 3} \dots (ii)$

From eq. (i) & (ii) $⟶ M = 23 gm / mol$

$M - 16 = 7$

Final answer: 7

Chemistry - Stoichiometry Question with Solution | TestHub

Doubts & Discussion