CHEM 51 West Los Angeles College Gas Density of Dichlorine Pentoxide Chemistry Ques

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CHEM 51

West Los Angeles College

CHEM

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1. A 10.00 gram mixture of NaHCO3 and Na2CO3 was heated and yielded 0.0357 moles of H2O and 0.1091 moles of CO2. The reactions are described by: NaHCO3 -----> Na2O + CO2 + H2O and Na2CO3 -----> Na2O + CO2. Calculate the volume of CO2 produced in each reaction at 733 Torr and 43oC. and the percentage composition of the mixture.

2. A 5.00 cubic yard tank contains 193 moles of nitrogen, 125 moles of oxygen, 145 moles of helium, and 169 moles of carbon dioxide at 220o F. Calculate the partial pressure of each gas in the mixture. 1 yard = 3 feet, 1 foot = 12 inches, and 1 inch = 2.54 cm.

3. In a photosynthesis reaction. 0.5796 grams of glucose was formed at a temperature of 37.0oC and at a pressure of 745 Torr. How many milliliters of oxygen were formed with the glucose and how many milliliters of carbon dioxide were reacted. The vapor pressure of water at 37.0oC is 50.0 Torr. The photosynthetic reaction is: 6CO2 + 6H2O -----> 6O2 + C6H12O6.

4. In the reaction of potassium fluoride with potassium chromate and sulfuric acid, the products are fluorine gas, potassium sulfate, chromium (III) sulfate and water. If 39.3 liters of fluorine gas was formed in this reaction and the gas was collected at -10.0oC and 725 Torr, how many grams of potassium chromate reacted?

5. In the reaction of hydrochloric acid and sodium hydroxide, 100.00 ml of 0.1234M hydrochloric acid reacts with 50.00 ml of 0.5432M sodium hydroxide. Will the final solution be more acidic or more basic? How many mole of excess acid or base will remain? What is the final (H+ ) and what is the final pH?

6. In the decomposition reaction of indium perchlorate, 25.0 grams is heated to produce oxygen and indium chloride. Write the balanced equation for the reaction and find the mole ratios of the reactants and products required to solve any stoichiometric problem; then find the volume of oxygen in liters formed at 37oC if the gas is collected over mercury under an atmospheric pressure of 720 torr.

7. If 30.0 ml of 0.150 M CaCl2 is added to 27.5 ml of 0.200 M AgNO3, how many grams of AgCl will be precipitated?

8a. What is the gas density of dichlorine pentoxide at 1500 Torr. and 270oC?

8b. If the gas density of a given compound is 3.75 grams/liter at 960 Torr and 55oC, what is the molar mass of the gas?

9. A large weather balloon is filled with 1500. moles of helium at 35.0oC and 1.076 atm. What is its final volume at altitude where the pressure of the atmosphere is 25.0 Torr and the final temperature is -350o F?

10a. 1.00 liter flask was evacuated and 5.00 grams of liquid NH3 was added to the flask, and the flask was sealed with a rubber stopper. If it takes 7.10 atm of pressure in the flask to blow out the rubber stopper, at what temperature will the stopper be blown out?

10b. When ammonium nitrite is heated it decomposes to give nitrogen gas and water. This property is used to inflate some tennis balls. Write a balanced equation for the reaction. Then calculate the quantity of ammonium nitrite needed to inflate 10 tennis balls each to a volume of 86.2 ml at 1.2 atm. and 22oC.

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Explanation & Answer

Hey buddy. Attached are the answers to your questions above. Feel free to ask any questions regarding the order :) Thank you!

1.
Given:
𝑚𝑁𝑎𝐻𝐶𝑂3 + 𝑚𝑁𝑎2𝐶𝑂3 = 10.00 𝑔
𝑛 𝑇𝑜𝑡𝑎𝑙 𝐻2𝑂 = 0.0357 𝑚𝑜𝑙
𝑛 𝑇𝑜𝑡𝑎𝑙 𝐶𝑂2 = 0.1091 𝑚𝑜𝑙
2𝑁𝑎𝐻𝐶𝑂3 → 𝑁𝑎2 𝑂 + 2𝐶𝑂2 + 𝐻2 𝑂
𝑁𝑎2 𝐶𝑂3 → 𝑁𝑎2 𝑂 + 𝐶𝑂2
Required: Volume of 𝐶𝑂2 in each reaction at 733 torr and 43℃
Solution:
𝑛 𝑇𝑜𝑡𝑎𝑙 𝐻2𝑂 = 𝑛𝐻2𝑂 𝑓𝑟𝑜𝑚 𝑁𝑎𝐻𝐶𝑂3
0.0357 𝑚𝑜𝑙 = 𝑚𝑁𝑎𝐻𝐶𝑂3 ×

1 𝑚𝑜𝑙 𝑁𝑎𝐻𝐶𝑂3
1 𝑚𝑜𝑙 𝐻2 𝑂
×
84.007 𝑔
2 𝑚𝑜𝑙 𝑁𝑎𝐻𝐶𝑂3

𝑚𝑁𝑎𝐻𝐶𝑂3 = 5.9980998 𝑔
𝑛 𝑇𝑜𝑡𝑎𝑙 𝐶𝑂2 = 𝑛𝐶𝑂2 𝑓𝑟𝑜𝑚 𝑁𝑎𝐻𝐶𝑂3 + 𝑛𝐶𝑂2 𝑓𝑟𝑜𝑚 𝑁𝑎2 𝐶𝑂3
0.1091 𝑚𝑜𝑙 = 𝑚𝑁𝑎𝐻𝐶𝑂3 ×

1 𝑚𝑜𝑙 𝑁𝑎𝐻𝐶𝑂3
2 𝑚𝑜𝑙 𝐶𝑂2
1 𝑚𝑜𝑙 𝑁𝑎2 𝐶𝑂3
1 𝑚𝑜𝑙 𝐶𝑂2
×
+ 𝑚𝑁𝑎2𝐶𝑂3 ×
×
84.007 𝑔
2 𝑚𝑜𝑙 𝑁𝑎𝐻𝐶𝑂3
105.9888 𝑔
1 𝑚𝑜𝑙 𝑁𝑎2 𝐶𝑂3

0.1091 𝑚𝑜𝑙 = 5.9980998 𝑔 ×

1 𝑚𝑜𝑙 𝑁𝑎𝐻𝐶𝑂3
2 𝑚𝑜𝑙 𝐶𝑂2
1 𝑚𝑜𝑙 𝑁𝑎2 𝐶𝑂3
1 𝑚𝑜𝑙 𝐶𝑂2
×
+ 𝑚𝑁𝑎2𝐶𝑂3 ×
×
84.007 𝑔
2 𝑚𝑜𝑙 𝑁𝑎𝐻𝐶𝑂3
105.9888 𝑔
1 𝑚𝑜𝑙 𝑁𝑎2 𝐶𝑂3
𝑚𝑁𝑎2 𝐶𝑂3 = 3.99577776 𝑔

𝑛𝐶𝑂2 𝑓𝑟𝑜𝑚 𝑁𝑎𝐻𝐶𝑂3 = 5.9980998 𝑔 ×

1 𝑚𝑜𝑙 𝑁𝑎𝐻𝐶𝑂3
2 𝑚𝑜𝑙 𝐶𝑂2
×
84.007 𝑔
2 𝑚𝑜𝑙 𝑁𝑎𝐻𝐶𝑂3

𝑛𝐶𝑂2 𝑓𝑟𝑜𝑚 𝑁𝑎𝐻𝐶𝑂3 = 0.0714 𝑚𝑜𝑙
𝑃𝑉 = 𝑛𝑅𝑇
733 𝑡𝑜𝑟𝑟 ×

1 𝑎𝑡𝑚
𝐿 ∙ 𝑎𝑡𝑚
× 𝑉 = 0.0714 𝑚𝑜𝑙 × 0.08205
× (43 + 273.15)𝐾
760 𝑡𝑜𝑟𝑟
𝑚𝑜𝑙 ∙ 𝐾
𝑉𝐶𝑂2 𝑓𝑟𝑜𝑚 𝑁𝑎𝐻𝐶𝑂3 = 1.920 𝐿

𝑛𝐶𝑂2 𝑓𝑟𝑜𝑚 𝑁𝑎2 𝐶𝑂3 = 3.99577776 𝑔 ×

1 𝑚𝑜𝑙 𝑁𝑎2 𝐶𝑂3
1 𝑚𝑜𝑙 𝐶𝑂2
×
105.9888 𝑔
1 𝑚𝑜𝑙 𝑁𝑎2 𝐶𝑂3

𝑛𝐶𝑂2 𝑓𝑟𝑜𝑚 𝑁𝑎2𝐶𝑂3 = 0.0377 𝑚𝑜𝑙
733 𝑡𝑜𝑟𝑟 ×

1 𝑎𝑡𝑚
𝐿 ∙ 𝑎𝑡𝑚
× 𝑉 = 0.0377 𝑚𝑜𝑙 × 0.08205
× (43 + 273.15)𝐾
760 𝑡𝑜𝑟𝑟
𝑚𝑜𝑙 ∙ 𝐾
𝑉𝐶𝑂2 𝑓𝑟𝑜𝑚 𝑁𝑎2 𝐶𝑂3 = 1.014 𝐿

%𝑁𝑎𝐻𝐶𝑂3 =

5.9980998 𝑔
× 100%
10 𝑔

%𝑁𝑎𝐻𝐶𝑂3 = 59.98%
%𝑁𝑎2 𝐶𝑂3 =

3.99577776 𝑔
× 100%
10 𝑔

%𝑁𝑎2 𝐶𝑂3 = 39.96%

2.
Given:
𝑉 = 5.00 𝑦𝑑 3
𝑛𝑁2 = 193 𝑚𝑜𝑙
𝑛𝑂2 = 125 𝑚𝑜𝑙
𝑛𝐻𝑒 = 145 𝑚𝑜𝑙
𝑛𝐶𝑂2 = 169 𝑚𝑜𝑙
𝑇 = 220℉ = 104.44℃ = 377.6 𝐾
Required: Partial Pressure of each gas
Solution:
3 𝑓𝑡 12 𝑖𝑛 2.54 𝑐𝑚 3
1𝐿
𝑃𝑇𝑜𝑡𝑎𝑙 × 5 𝑦𝑑 3 × (
×
×
) ×
1 𝑦𝑑 1 𝑓𝑡
1 𝑖𝑛
1000 𝑐𝑚3
= (193 + 125 + 145 + 169)𝑚𝑜𝑙 × 0.08205
𝑃𝑇𝑜𝑡𝑎𝑙 = 5.12 𝑎𝑡𝑚
𝑛 𝑇 = 193 + 125 + 145 + 169
𝑛 𝑇 = 632 𝑚𝑜𝑙
𝑃𝑖 = 𝑃𝑇 ×

𝑛𝑖
𝑛𝑇

𝑃𝑁2 = 5.12 𝑎𝑡𝑚 ×

193 𝑚𝑜𝑙
632 𝑚𝑜𝑙

𝑃𝑁2 = 1.56 𝑎𝑡𝑚
𝑃𝑂2 = 5.12 𝑎𝑡𝑚 ×

125 𝑚𝑜𝑙
632 𝑚𝑜𝑙

𝑃𝑂2 = 1.01 𝑎𝑡𝑚
𝑃𝐻𝑒 = 5.12 𝑎𝑡𝑚 ×

145 𝑚𝑜𝑙
632 𝑚𝑜𝑙

𝑃𝐻𝑒 = 1.17 𝑎𝑡𝑚
𝑃𝐶𝑂2 = 5.12 𝑎𝑡𝑚 ×

169 𝑚𝑜𝑙
632 𝑚𝑜𝑙

𝑃𝐶𝑂2 = 1.37 𝑎𝑡𝑚

𝐿 ∙ 𝑎𝑡𝑚
× 377.6 𝐾
𝑚𝑜𝑙 ∙ 𝐾

3.
Given:
𝑚𝐶6 𝐻12𝑂6 = 0.5796 𝑔
𝑇 = 37.0℃
𝑃𝑤𝑒𝑡 𝑔𝑎𝑠 = 745 𝑡𝑜𝑟𝑟
𝑃𝐻2𝑂 = 50.0 𝑡𝑜𝑟𝑟
6𝐶𝑂2 + 6𝐻2 𝑂 → 6𝑂2 + 𝐶6 𝐻12 𝑂6
Required: Volume of 𝑂2 formed (mL) and Volume of 𝐶𝑂2 reacted (mL)
Solution:
𝑛𝑂2 formed = 0.5796 𝑔 𝐶6 𝐻12 𝑂6 ×

1 𝑚𝑜𝑙 𝐶6 𝐻12 𝑂6
6 𝑚𝑜𝑙 𝑂2
×
180.156 𝑔
1 𝑚𝑜𝑙 𝐶6 𝐻12 𝑂6

�...


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