Ch 3 Q74 and Ch4 Q70

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American Military University *

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Chemistry

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Jun 21, 2024

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Chapter 3, Q74: The level of mercury in a stream was suspected to be above the minimum considered safe (1 part per billion by weight). An analysis indicated that the concentration was 0.68 parts per billion. Assume a density of 1.0 g/mL and calculate the molarity of mercury in the stream. We know that the concentration of mercury is 0.68 ppb and that the density is 1.0 g/ mL so 10 to the 9 th power mL will hold 0.68 mL of mercury and 0.68 mL of mercury would equate to .68 g. Before calculating molarity, there needs to be a conversion of concentration from micrograms per liter to grams per liter. Given 0.68 ppb, and that 1 ppb is equivalent to 1 microgram/ L, 0.68 ppb equals 0.68 microgram/ L. Converting micrograms/L to g/L, 1 microgram is equal to 1 x 10 to the negative 6 th power grams which shows that 0.68 micrograms/L = 0.68 x 10 to the negative 6 th power g/L which also equates to 6.8 x 10 to the negative 7 th g/L Now for calculating the molarity, the molar mass of mercury is equal to 201 g/ mol. Using the formula of molarity = 6.8 x 10 to the negative 7 th power g/L divided by 201 g/ mol, this will give us the answer: 3.39 x 10 to the negative 9 th power Chapter 4, Q70: Outline the steps needed to determine the limiting reactant when 30.0 g of propane, C 3 H 8 , is burned with 75.0 g of oxygen. Determine the limiting reactant. With the word ‘burned’ in the problem, we know this is a combustion reaction which will produce CO 2 and H 2 O. The balanced chemical equation for this problem will be C 3 H 8 + 5O 2 yields 3CO 2 + 4H 2 O. Now we’ll calculate the molar masses by multiplying by the atomic masses of carbon (C) and hydrogen (H), starting with propane (C 3 H 8 ): 3 x 12.01 C + 8 x 1.01 H, which gives 36.03 + 8.08 that equals to 44.11 g/mol . And then oxygen (O 2 ): 2 x 16.00 that equals to 32 g/mol. Moving on, convert to moles of propane and oxygen. C 3 H 8 holds 30g that we will divide by the 44.11 g/mol this equals 0.68 mol. O 2 holds 75g that we will divide by 32 g/mol which equals 2.34 mol. We now know what we have and now it’s time to solve for what we need to find our limiting reactant. We have 30 g of C 3 H 8 and 75 g of O 2 so let’s calculate the amount of O 2 needed to react with the C 3 H 8 that we have. To find this we will multiply 0.68 mol of propane by 5 (mol of O 2 / mol of C 3 H 8 ) which gives us 3.4 mol. Once converting moles of O 2 to grams by taking the molar mass (32 g/mol) and multiplying it by the 3.4 mol that was just solved for, this shows us that we need roughly 109 grams of O 2 to react with the 30g of C 3 H 8 . Since 109 g is needed, this tells us that: Oxygen is the limiting reactant

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