Determine the mass (in grams) of Al2(CO3)3 that contains 8.99 x 10^23 carbonate ions

Question

Question

science chemistry

Answer 1

Answer #1

1 mole of Al₂(CO₃)₃ = 2 moles of Al³⁺ ions.

1 mole of Al₂(CO₃)₃ = 3 moles of CO₃^2- ions.

given that sample contains 8.99 * 10^23 carbonate ions so

number of moles of carbonate ions = 8.99 * 10^23 / 6.022 * 10^23 = 1.492 moles

number of moles of the sample = number of moles of carbonate ion / 3

= 1.492/3 = 0.497 moles

molar mass of the sample = 2 * (26.98) + 3 * (12) + 9 * 16 = 233.98 g/mole

so mass of 0.497 moles of the sample = molar mass * number of moles

= 233.98 * 0.497 g= 116.43 g

Answer 2

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