HW8 - Lin (cgl286) Homework 8 Lyon (52370) 1 This print-out...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Lin (cgl286) Homework 8 Lyon (52370) 1 This print-out should have 35 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. This HW assignment is due Wednesday, April 7, by 11PM. 001 10.0 points An element X combines with oxygen to form a compound of formula XO 2 . If 24.0 grams of element X combines with exactly 16.0 grams of O to form this compound, what is the atomic weight of element X? 1. 24.0 amu 2. 48.0 amu correct 3. 284 amu 4. 12.0 amu 5. 16.0 amu Explanation: m X = 24.0 g m O = 16.0 g We know from the formula XO 2 that one mole of X combines with exactly 2 moles of oxygen atoms. We can convert 16 grams of oxygen to moles using the atomic weight of oxygen from the periodic table: ? mol O = 16 g O 1 mol O 16 g O = 1 mol O 1 mol O would combine with exactly 1 / 2 mol of X: ? mol X = 1 mol O 1 mol X 2 mol O = 0 . 5 mol X We were given that 16 g of O combines exactly with 24 g X. We calculated that 16 g of O (1 mol O) would combine with 0 . 5 mol X. From this we can conclude that 0 . 5 mol X has a mass of 24 g. We can calculate the grams per mole: ? molar mass X = 24 g X . 5 mol X = 48 g X mol X The molar mass of an element in g/mol is numerically equal to the atomic mass in amu. Therefore, the atomic mass of X is 48 amu. 002 10.0 points 27 g of Al reacts with 48 g of S to produce Al 2 S 3 , the only product, with no reactant left over. How much Al 2 S 3 would be produced from 78 g of Al and 101 g of S? 1. 158 g correct 2. 179 g 3. 281 g 4. 122 g 5. 217 g 6. 27 g Explanation: m ini , Al = 27 g m ini , S = 48 g m Al = 78 g m S = 101 g By the law of conservation of mass, 27 g of Al and 48 g of S would produce 75 g of Al 2 S 3 , if no Al and S are remaining at the end of the reaction. By the law of definite proportions then, the mass ratio of product Al 2 S 3 to either reactant (Al or S) is unchanging. However, the mass ratio of Al to S is different in the reaction at hand than in the ideal reaction, so one of the reactants must be limiting. If we assume that Al is limiting, the following relationship is true: 75 g Al 2 S 3 27 g Al = x g Al 2 S 3 78 g Al x = 217 g Al 2 S 3 Then if we assume that S is limiting, 75 g Al 2 S 3 48 g S = y g Al 2 S 3 101 g S y = 158 g Al 2 S 3 . Since the number of grams of Al 2 S 3 is smaller when we assume S is the limiting Lin (cgl286) Homework 8 Lyon (52370) 2 reagent, S really is the limiting reagent and only 158 g of Al 2 S 3 is formed. 003 10.0 points Consider the reaction CaCN 2 + 3 H 2 O CaCO 3 + 2 NH 3 , which has a 45.6% yield. How much H 2 O is needed to produce 16.3 moles of CaCO 3 ? 1. 107 g 2. 401 g 3. 48.9 g 4. 22.3 g 5. 1930 g correct 6. 880 g Explanation: Due to the % yield, the 16.3 moles of CaCO 3 is the ACTUAL yield. We must find the THE- ORETICAL yield (which is always greater): (16 . 3 mol CaCO 3 ) 100% 45 . 6% = 35 . 7456 mol CaCO 3 Find the moles of water needed: (35 . 7456 mol CaCO 3 ) 3 mol H...
View Full Document

This note was uploaded on 10/26/2010 for the course CH 52370 taught by Professor Lyon during the Spring '09 term at University of Texas at Austin.

Page1 / 11

HW8 - Lin (cgl286) Homework 8 Lyon (52370) 1 This print-out...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online