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2 labs and 1 assignment needs to be done as per instructions in the attachments.
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module 5 assignment
name:
Chemical Equations: show work for credit
1.
How many moles of NO2 forms when 0.356 moles of N2O5 decomposes?:
2N2O5(g) → 4 NO2(g) + O2(g)
.356 g mol N2O5 x 4 mol NO2/2mol N2O5 = 0.712 mol NO2
page 1 of 3
Properties of Liquids and Solids
Instructions: Answer the following questions using the following information:
∆Hfus = 6.02 kJ/mol;
∆Hvap= 40.7 kJ/mol;
2. For the following reaction, calculate how many moles of each product are formed when 0.356
moles of PbS completely react. Assume there is an excess of oxygen.
2PbS(s) + 3O2(g) → 2PbO(s) + 2SO2(g)
This will make 0.356 moles of PbO and 0.356 moles of SO2.
specific heat of water is 4.184 J/g∙˚C;
specific heat of ice is 2.06 J/g∙˚C;
specific heat of water vapor is 2.03 J/g∙˚C.
6. How much heat is required to vaporize 25 g of water at 100˚C?
3. For the following reaction, calculate how many grams of each product are formed when 4.05 g of
water is used.
2H2O(g) → 2H2(g) + O2(g)
4.05g of water = 0.225 moles of water.
0.225 moles of water produces 0.225 moles of H2 = 0.45 g of H2
0.225 moles of water produces 0.1125 moles of O2 = 3.6 g of O2.
4. Determine the theoretical yield of P2O5, when 3.07 g of P reacts with 6.09 g of oxygen in the
following chemical equation:
4P + 5O2 → 2P2O5
O2 required: (3.07g P / 31g P) x (5/4) moles of O2 x (32g O2) = 3.96 g of O2
Limiting Reagent = P. So, P2O5 theoretical yield = (3.07g P / 31g P) x (2/4) x (142 g
P2O5) = 7.03 g
5. Determine the percent yield of the following reaction when 2.80 g of P reacts with excess oxygen.
The actual yield of this reaction is determined to by 3.89 g of P2O5.
4P + 5O2 → 2P2O5
O2 required: (2.80g P / 31g P) x (5/4) moles of O2 x (32g O2) = 3.61g of O2
Theoretical yield = 2.80 + 3.61 g = 6.41g
Percentage yield = (3.89/6.41) x 100 = 60.7%
25g of water = 25/18 = 1.389 moles
Q = ∆Hvap x 1.389 moles = 40.7 x 1.389 = 56.5 kJ
7. How much heat is required to convert 25 g of ice at -4.0 ˚C to water vapor at 105 ˚C (report
your answer to three significant figures)?
25g ice = 1.389 moles
Heat required = (2.06 x 25 x 4) + (1.389 x 6020) + (4.184 x 25 x 100) + (1.389 x 40700) +
(2.03 x 25 x 5) = 75.8 kJ
8. An ice cube at 0.00 ˚C with a mass of 8.32 g is placed into 55 g of water, initially at 25 ˚C. If
no heat is lost to the surroundings, what is the final temperature of the entire water sample after
all the ice is melted (report your answer to three significant figures)?
8.32g ice = 8.32/18 = 0.462 moles
Let, final temperature is T.
(6020 x 0.462) + (4.184 x 8.32 x T) = (4.184 x 55 x (25 - T))
So, T = 11.2 ˚C
module 5 assignment
name:
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