Solution Stoichiometry Key

Solution Stoichiometry CHEMISTRY 110 Name________________________ last first 1] How many grams of calcium phosphate can be produced from the reaction of 2.50 L of 0.250 M Calcium chloride with and excess of phosphoric acid? Calcium chloride + phosphoric acid --> calcium phosphate + hydrochloric acid 3CaCl2 + 2H3PO4 ---> Ca3(PO4)2 + 6HCl 2.50L CaCl2 X. 0.250L CaCl2 1mol Ca3(PO4)2 310.0g Ca3(PO4)2 X. X. = 1mol 3mol CaCl2 1mol Answer ____________________ 64.6g Ca3(PO4)2 2] How many milliliters of 1.50 M Nitric acid is required to react with 100.0 g of cuprous oxide 14 HNO3 + 3Cu2O ---> 6 Cu(NO3)2 + 2 NO + 7H2O 1mol Cu2O 14mol HNO3 1000ml HNO3 X 3mol Cu O X 1.50mol HNO = 2.18 x 103ml HNO3 100.0g Cu2O X 143.1g 2 3 Answer _________________________ 3] 60.5 mL of HNO3 are required to react with 25.0 mL of a 1.00 M Barium hydroxide solution: 2 HNO3(aq) + Ba(OH)2(aq) --> 2 H2O(s) + Ba(NO3)2(aq) (BALANCED) Find the Molarity of the nitric acid solution 1.00mol Ba(OH)2 2mol HNO3 1L X. X 25.0ml Ba(OH)2 X 3 1L 1mol Ba(OH)2 = 0.0500mol HNO3 10 ml 0.0500 mol HNO3 = 0.826 M 0.0605 L soln Answer _________________________ 4] For the following equation determine which reactant is the limiting reactant and which reactant is in excess. The amounts of reagent used are shown. Show calculations to support your choices 3Fe + 4H2O ----> Fe3O4 + 4H2 40.0 g 16.0g MHNO3 = 1molFe 1mol Fe3O4 40.0g Fe X 55.8g X 3molFe = 0.239 mol Fe3O4 16.0g H2O X 1molH2O 1mol Fe3O4 18.0g X 4mol H2O = 0.222 mol Fe3O4 <==== amount made The excess reactant is ______Fe______ The limiting reactant is ____H2O_____ 5] 35.5 g of silver nitrite is reacted with 35.5 grams of sodium sulfide which produces silver sulfide and sodium nitrite. a. Write and balance the equation 2AgNO2 + Na2S ---> Ag2S + 2NaNO2 b.. Calculate the number of grams of silver sulfide produced. 1mol AgNO2 1mol Ag2S 35.5g AgNO2 X X 2mol AgNO2 = 0.115 mol Ag2S<==== amount made 153.9g 35.5g Na2S X. 1mol Na2S 1mol Ag2S X 78.0g 1mol Na2S = 0.455 mol Ag2S 247.8g Ag2S = 28.5g Ag2S 1mol Answer _________________________ 0.115 mol mol Ag2S X c. How many grams of silver nitrite will remain at the end of the reaction? Answer _______0 g_____ The AgNO2 is the limiting reactant d. How many grams of sodium sulfide will remain at the end of the reaction? 1mol Na2S 78.0 g Na2S 0.115 mol Ag2S X 1 mol Ag S X = 8.97g Na2S used 1mol 2 Excess Na2S = 35.5g - 8.97g = 26.5g Na2S Answer _________________________ 6] Calculate the grams of silver chloride produced from 10.00 ml of 10.0M magnesium chloride with 100.0 ml of 2.20 M silver nitrate 2AgNO3 + MgCl2-->Mg(NO3)2(s) + 2AgCl(aq) 0.01000 L MgCl2 X 10.0mol MgCl2 2mol AgCl X 1mol MgCl = 0.200 mol AgCl<==== amount made 1L 2 0.1000 L AgNO3 X 2.20molAgNO3 2mol AgCl X 2mol AgNO = 0.220 mol AgCl 1L 3 143.3g AgCl = 28.7g AgCl 1mol Answer _________________________ 7] Aluminum reacts with oxygen to form aluminum oxide: 4Al + 3O2 ---> 2Al2O3 (balanced) If 75.0g of Al and 200.0 g of oxygen are reacted, and 75.0 g of aluminum oxide is produced, what is the percent yield for the reaction? 0.200 mol AgCl X 1mol Al 2mol Al2O3 75.0g Al X 27.0g X 4mol Al = 1.39 mol Al2O3<==== amount made 1mol O2 2mol Al2O3 200.0 g O2 X 32.0g X 3mol O = 4.17 mol Al2O3 2 1.39 mol Al2O3 X 102.0g Al2O3 = 141.8 g Al2O3 1mol 75.0g % Yield = 141.8g (100) = 52.9% Answer _________________________ 8] . According to the following reaction:..... 2 Cu(s) + O2(g) ----> + 2 CuO(s) a. If the percentage yield is 96.7% how many grams of CuO will be produced from 13.4 g of Cu? 1mol Cu 2mol CuO 79.5gCuO 96.7g Actual 13.4 g Cu X 63.5g X 2mol Cu X 1mol X 100g Theo. = 16.2 g CuO Answer _________________________ b..How many grams of Cu must you use to produce 5.00 x 1013mg CuO? 10-3 1mol CuO 2mol Cu 63.5g Cu 100g 5.00 x 1013mg CuO X 1mg X 79.5g X 2mol CuO X 1mol X 96.7g = 4.13 x 1010g Answer ___________________