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

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.