Using the average atomic masses given inside the front cover of this book, calcu

Question

Using the average atomic masses given inside the front cover of this book, calculate how many moles of each substance the following masses represent
(b) 2.10 mg of gold(III) chloride, AuCl3

Using the average atomic masses given inside the front cover of this book, calculate the mass in grams of each of the following samples.
(b) 4.45 mmol of ammonia, NH3 (1 mmol = 1/1000 mol)

For each of the following unbalanced equations, calculate how many grams of each product would be produced by complete reaction of 12.3 g of the first reactant.

(a) TiBr4(g) + H2(g) Ti(s) + HBr(g)

(b) BaCl2(aq) + AgNO3(aq) AgCl(s) + Ba(NO3)2(aq)

Explanation / Answer

b)

No of mol of AuCl3 = 2.1*10^(-3) / 303.3256

      = 6.92*10^-6 mol

c) 4.45*10^(-3) mol NH3

NH3 = 4.45*10^(-3)*17 = 0.07565 grams

a) TiBr4(g) + 2H2(g) Ti(s) + 4HBr(g)

No of mol of TiBr4 = 12.3/367.48   = 0.0334mol

mass of Ti = 0.0334*47.867 = 1.6 grams

mass of HBr = 4*0.0334*80.91 = 10.8 grams

(b) BaCl2(aq) + 2AgNO3(aq) 2AgCl(s) + Ba(NO3)2(aq)

No of mol of BaCl2 = 12.3/208.2330 = 0.06 mol

mass of AgCl = 0.06*143.32*2 = 17.2 grams

mass of Ba(NO3)2 = 261.37*0.06 = 15.68 GRAMS

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