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Unformatted text preview: Homework Questions Week 8 SOLUTIONS 22 Marks total 1. 75mL of 0.25M hydrochloric acid is added to 5.00g of calcium carbonate. a. Write a balanced equation for this reaction b. Find the number of moles of each reactant c. Find the limiting reactant d. Calculate the theoretical mass of calcium chloride solid which will be produced if you completely evaporate the remaining solution. 2HCl + CaCO3 .> CaCl2 + H2O + CO2 b. 0.01875 mol of HCl is present M CaCO3 = 100g n CaCO3 = 0.05 The acid is the limiting reagent 0.01875/2 mol of NaCl will form = 0.55g of NaCl will be produced (6 marks) 2. Magnesium 2.5g is burnt in 2.5g of carbon dioxide gas. What mass of carbon should be displaced during this reaction a. Write a balanced chemical equation for this change b. Calculate the number of moles of both reactant c. Find the volume of the carbon dioxide at STP d. Calculate the mass of carbon produced a) 2Mg + CO2 à 2MgO + C b) 2.5g of Mg = 2.5/24 mol = 0.104mol 2.5g of CO2 = 2.5/44 mol = 0.0568mol this would require 0.113mol of Mg for complete reaction Mg is limiting reagent c) 0.0568mol of CO2 is present this equals = 1.27L of gas d) The mass of C produced depends upon the amount of LR. 0.104/2 = 0.052mol = 0.624g of C produced (6marks) CSC Chemistry 11 T1 W8 SOLUTIONS 3 Use the foll owing equation to solve the question below c1V1 = c2V2 c1 = concentration of concentrated acid, V1 = volume of concentrated acid, c2 = concentration of dil uted solution, V2 = volume of diluted solution What volume of concentrated hydrochloric acid (12M) is required to produce 500mL of 0.25M hydrochl oric acid (diluted acid)? Substitute concentrated acid values (1) and changes (2) 12M X V1 = 0.25M x 0.5L 12M x V1 = 0.125 V1 = 0.125/12 = 0.0104L Therefore 0.0104 x 1000mL is required 10.4mL of conc acid needs to be made up to 500mL with water (3 marks) 3. Research the ideal gas equation. a. What is the equation b. List the variables c. State the correct SI units for these variables d. Rearrange the equation so that V (volume) is the subject Application for the experts……? e. What is the volume of 0.45mol of an ideal gas at 35 degrees C and at 98kPa pressure (express V in L) pV = nRT 3 B+c) p = pressure (Pa) V = volume (m ) n = number of mol R = gas constant T = temp (K) d) V = nRT/p CSC Chemistry 11 T1 W8 SOLUTIONS (4 marks, 2 for parts B+C) e) V = 0.45 x 8.314 x 308K/ 98000 Pa (answer in metres cube) Ans = 0.0112m3 of gas or 11.76L of ideal gas (3 marks) CSC Chemistry 11 T1 W8 SOLUTIONS ...
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This note was uploaded on 09/27/2010 for the course 56 1121 taught by Professor Khjdf during the Spring '10 term at Mackenzie.
- Spring '10