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Mole calculations

Mole calculations. How many mol of BaCl 2 are contained in 2.74 g of BaCl 2 ? M (Ba) = 137.33 g mol –1 , M (Cl) = 35.5 g mol –1. 2.74 g. Work out the molar mass of BaCl 2 . M (BaCl 2 ) = M (Ba) + 2  M (Cl) = 137.33 + (2  35.5) = 208.33 g mol –1. 208.33 g mol –1.

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Mole calculations

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  1. Mole calculations

  2. How many mol of BaCl2 are contained in 2.74 g of BaCl2? M(Ba) = 137.33 g mol–1, M(Cl) = 35.5 g mol–1. 2.74 g Work out the molar mass of BaCl2. M(BaCl2) = M(Ba) + 2  M(Cl) = 137.33 + (2  35.5) = 208.33 g mol–1 208.33 g mol–1 Calculate the amount of BaCl2 present.

  3. What is the mass of 2.66  10–3 mol of BaSO4? M(BaSO4) = 233.4 g mol–1 2.66  10–3 mol 233.4 g mol–1 m(BaSO4) = nM = 2.66  10–3 mol 233.4 g mol–1 = 0.621 g

  4. How many moles of Na+ are present in 6.48 g of Na2CO3? M(Na2CO3) = 106 g mol–1. 6.48 g 106 g mol–1 First work out how many moles of Na2CO3 there are. 6.08  10–2 mol In each mol of Na2CO3 there are 2 moles of Na+. 2 2 so n(Na+) = 2  n(Na2CO3) = 2  6.08  10–2 mol = 0.122 mol

  5. How many mol of NO3– in 14.00 g of Fe(NO3)3.9H2O? Ar: Fe = 55.85; N = 14.01; O = 16.00; H = 1.01 14.00 g The ‘.9H2O’ part means that there are 9 moles of water trapped with each mole of iron(III) nitrate. Think of the dot as a plus sign. M(Fe(NO3)3.9H2O = Fe + (3 (N + O3) ) + ( 9 (H2 + O) ) 55.85 + (3  (14.01 + (3 16.00) + (9 (2 1.01) + 16.00)) = 404.06 g mol–1 404.06 g mol–1

  6. Now calculate the number of moles of NO3– in 3.465  10–2 mol of Fe(NO3)3.9H2O. 3.465  10–2 mol In each mol of Fe(NO3)3.9H2O there are 3 mol of NO3–. 3 3 n(NO3–) = n(Fe(NO3)3.9H2O)  3 = 3.465  10–2 mol  3 = 0.1039 mol

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