What is the amount (in mole) of sodium trioxocarbonate (IV) in 5.3 g of the compound? (Na2CO3 = 106)

Assessment: WAEC SSCE - Chemistry - 1991 (Objective) Subject: Chemistry

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What is the amount (in mole) of sodium trioxocarbonate (IV) in 5.3 g of the compound? (Na2CO3 = 106)
Answer Details
The molar mass of sodium trioxocarbonate (IV) is 2(23) + 12 + 3(16) = 106 g/mol (by adding the atomic masses of the elements present in one mole of the compound). To find the amount (in moles) of the compound in 5.3 g, we need to divide the given mass by the molar mass: Amount of Na2CO3 = Mass of Na2CO3 / Molar mass of Na2CO3 Amount of Na2CO3 = 5.3 g / 106 g/mol Amount of Na2CO3 = 0.05 mol Therefore, the amount (in moles) of sodium trioxocarbonate (IV) in 5.3 g of the compound is 0.05. The answer is.
Read lesson note on Chemical Combination (JAMB)
Read lesson note on Stoichiometry And Chemical Reactions (WAEC)
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