Practice with Molar Mass Problems
:Question #1 How many moles of H 2 are in 100 g of H 2 ? # mol H 2 = 100 g H 2 x 1 mol H g H 2 = 49.5 mol H 2
Question #2 300 g of CuSO 4 is needed in an experiment. How many moles does this represent? # mol CuSO 4 = 300 g CuSO 4 x 1 mol CuSO g CuSO 4 = 1.88 mol CuSO 4
Question #3 A chemical reaction requires moles of silver chloride. How many grams is this? # g AgCl = mol AgClx g AgCl 1 mol AgCl = 3408 g AgCl
73 g H 2 O Question #4 How many molecules are in 73 grams H 2 O? # H 2 O molecules = x 1 mol H 2 O g H 2 O = 2.4 x molecules H 2 O x 6.02x10 23 molecules 1 mol H 2 O
Question #5 255 g of calcium phosphate are produced in a chemical reaction. How many moles of calcium phosphate does this represent? # mol Ca 3 (PO 4 ) 2 = 255 g Ca 3 (PO 4 ) 2 x 1 mol Ca 3 (PO 4 ) g Ca 3 (PO 4 ) 2 = mol Ca 3 (PO 4 ) 2
Question #6 According to the equation 2H 2 + O 2 2H 2 O, how many grams of H 2 O would be produced if 7.35 mol of O 2 is used up? (hint: you will need two conversion factors – 1 from the balanced equation and 1 from a molar mass) 2 mol H 2 O 1 mol O 2 x # g H 2 O= 7.35 mol O g H 2 O = g H 2 O 1 mol H 2 O x
And you thought Chemistry was bad with factor-label problems!!??
Complete the following chart (#7 - #13): FormulaMolar mass (g/mol) Mass (g) Moles (mol) FeSO (NH 4 ) 2 CO 3 2 SnO 2 50 Sb 2 O NaClO Mg(IO 3 ) CoCl 2.H 2 O332
Complete the following chart (answers): CoCl 2.H 2 O Mg(IO 3 ) NaClO Sb 2 O SnO (NH 4 ) 2 CO FeSO 4 Moles (mol) Mass (g) Molar mass (g/mol) Formula
Assignment 14. AgCl = g/mol #g = 2 mol x g/mol = g (2) 15. H 2 = g/mol #mol = 100 g x mol/2.016 g = 49.6 mol (2) 16. CuSO 4 = g/mol #mol= 300 g x mol/ g=1.879 mol (2) 17. KClO = g/mol #mol = 250 g x mol/90.55 g = 2.76 mol (2)
Extra Review Mass-Mole-Molecules: Determine the number of molecules in 73 g of water 73 g H 2 O # H 2 O molecules = x 1 mol H 2 O g H 2 O = 2.4 x molecules H 2 O x 6.02x10 23 molecules 1 mol H 2 O
Try this one: Calculate the mass in grams of iodine required to react completely with 0.50 moles of aluminum. Al + I 2 AlI 3 2 Al + 3 I 2 2 AlI 3 = 190 g I mol Al x 3 mol I 2 2 mol Al x g I 2 1 mol I 2
0.50 g Al Try this one: Calculate the mass in grams of iodine required to react completely with 0.50 g of aluminum. Al + I 2 AlI 3 2 Al + 3 I 2 2 AlI 3 = 7.1 g I 2 x 3 mol I 2 2 mol Al x g I 2 1 mol I 2 x 1 mol Al g Al