How many grams of residue is obtained by heating 2.76 g of silver carbonate

Question

How many grams of residue is obtained by heating 2.76 g of silver carbonate?

(Given: Molar mass of C, O and Ag are 12, 16 and 108 g ${mol}^{- 1}$ respectively) [2026]

Smart Achievers · Accepted Answer

(2)

${mol}^{- 1}$

$= (2 \times 108 + 12 + 3 \times 16) {g mol}^{- 1} = 276 {g mol}^{- 1}$

$Moles of silver carbonate (n_{A g_{2} C O_{3}})$

$= \frac{Mass of A g_{2} C O_{3}}{Molar mass of A g_{2} C O_{3}}$

$= (\frac{2.76}{276}) mol = 0.01 mol$

$A g_{2} C O_{3} \overset{Δ}{\to} 2 A g + \frac{1}{2} O_{2} + C O_{2}$

$As per reaction stoichiometry:$

$Moles of Ag = 2 \times moles of A g_{2} C O_{3}$

$Moles of Ag = 2 \times 0.01 mol = 0.02 mol$

$Mass of Ag = moles of Ag \times molar mass of Ag$

$= 0.02 \times 108 g = 2.16 g$

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