Question

1) The combustion of a sample of butane, C.H(lighter fluid), produced 2.46 grams of water. 2 C4H10 + 1302 -------> 8CO2 + 10H20 (a) How many moles of water formed? (b) How many moles of butane burned? (c) How many grams of butane burned? (d) How much oxygen was used up in moles? (e) How much oxygen was used up in grams? 2) Using the following equation: 2 NaOH + H2SO4 → 2 H2O + Na2SO4 How many grams of sodium sulfate will be formed if you start with 200 grams of sodium hydroxide and you have an excess of sulfuric acid? 3) Using the following equation: Pb(SO4)2 + 4 LINO3 → Pb(NO3)2 + 2 Li2SO4 How many grams of lithium nitrate will be needed to make 250 grams of lithium sulfate, assuming that you have an adequate amount of lead (IV) sulfate to do the reaction?

Answer 1

1) . 2.46 a) Moles of water formed -- g 0.137 mol b) 10 mol H2O = 2 mol CHIO 10.137 mol Ho - 2x0.137 mot con mol CHIO = 0.0274 mol 10 i. Moles of butame burned = 10.0274 moll c) Grams ob butane burned = 0.0274 mol x 58 gimol = 1.5892 g] d 10 mol H, 0 = ,: 0.137 mol H, 0 = 13 mol Oz 13x0.137 mol Q = 0.1781 mol Oz 10 i Amount of oxygen used = (0.1781 mol e) Giams of oxygen used = 0.1781 mol x 32 g/mol = [5,6992 g paget

4 According to reaction — 2x40 g Naolt produces 142 g Najsou 2009 Naom 142X200 g Na, soy 2x40 g Naz Soy = [ 355 g Nasou) 3) 2x 110 g Li,S0y is produced from 4x69 g Linoz :: 250 g Li, sou " 4x69x250 g Lines 2x110 = 313.64 g LINO3 page-2)