Chemistry: The Central Science (13th Edition)

Published by Prentice Hall
ISBN 10: 0321910419
ISBN 13: 978-0-32191-041-7

Chapter 10 - Gases - Exercises: 10.55

Answer

4.11 x 10^-9 grams

Work Step by Step

Since we have the volume, the partial pressure of O2 and the temperature, we can use the ideal gas law to find the moles of O2 that are in the enclosure. We have 3.5 x 10^-6 torr, which we convert to atm: 3.5 x 10^-6 torr * (1atm/760 torr) = 4.61 x 10^-9 atm Then we convert the 27C to Kelvin by adding 273, which is 300K Then we an use PV = nRT or n = PV/RT = (4.61 x 10^-9 atm * 0.452L) / (0.0821 * 300K) = 8.45 x 10^-11 moles Then we can use the molecular formula to see how many moles of Magnesium will react. Since 2 moles of Magnesium are required for one mole of O2, we will have: 2* 8.45 x 10^-11 = 1.69 x 10^-10 moles of Magnesium Then to convert to grams, we have the molar mass is 24.305 g/mol so 1.69 x 10^-10 moles * (24.305 g/ 1mol) = 4.11 x 10^-9 grams
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