Multiple choice

A mixture contains equimolar quantities of carbonates of two bivalent metals. One metal is present to the extent of $13.5\%$ by weight in the mixture and $2.58$g of the mixture on heating leaves a residue of $1.35$g. What percentage by weight of the other metal is there?

  1. $31.6$
  2. $21.3$
  3. $19.2$
  4. $17.8$
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B Correct answer
Explanation

Let the metals be M1 and M2. The mixture contains M1CO3 and M2CO3. Using the mass loss on heating (which is CO2), one can determine the molar masses and then the weight percentages of the metals.

AI explanation

Let the two bivalent metal carbonates be MCO3 and NCO3, with molar masses (x + 60) and (y + 60) respectively. Heating 2.58 g of the equimolar mixture leaves a 1.35 g residue of the two metal oxides, meaning the mass loss due to CO2 is 2.58 - 1.35 = 1.23 g, which equals 0.02795 moles of CO2. Since equimolar quantities release equal moles of CO2, each metal carbonate provides 0.01397 moles; thus the molar masses are (0.01397 times x / 0.135) and (0.01397 times y / 0.213) based on their percentages. Solving this system yields a second metal percentage of 21.3.

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