Chapter 2 Problems
20 original chapter problems from Chemistry I, Chapter 2. Review quizzes remain in their original lessons.
Original reading → · Source answers and worked examples →
-
Calculate the molecular weight for: C12H9NO2
-
1 SO42- ion = ___ g SO42- = ___ amu SO42-
-
How many moles of InBr2 are present in 50.2 g of InBr2 (MW = 274.6)? What is the mass percent of In in this compound?
-
Balance the following: C3H6O2 + O2 → CO2 + H2O
-
When the following equation is balanced, what is the sum of its coefficients: Na3PO4 + FeCl2 → NaCl + Fe3(PO4)2
-
How many grams of KClO3 would be required to produce 10.0 g of O2: KClO3 → KCl + O2 (unbalanced)?
-
If a 6.00 g sample of 97% pure H2 was reacted with 50.0g CO, how many moles of CH3OH are produced: H2 + CO → CH3OH (unbalanced)
-
If a 6.00 g sample of 97% pure H2 was reacted with 50.0g CO, how many moles of CH3OH are produced: H2 + CO → CH3OH (unbalanced)
Using the same equation as above, how many grams of CO are needed to completely react with 10.0 g of H2?
-
AP edition reviewed standalone replacement.
Reviewed standalone methanol replacement:react5.0gH₂ with3.00g of a96.5%-by-massCO sample. Assume the impurity is inert and useCO+2H₂→CH₃OH, M(H₂)=2.016andM(CO)=28.01g mol⁻¹. Identify the limiting reactant and theoretical methanol amount.
-
If a 6.00 g sample of 97% pure H2 was reacted with 50.0g CO, how many moles of CH3OH are produced: H2 + CO → CH3OH (unbalanced)
Using the same equation as above, if 5.0 g of H2 are reacted with 10.00 g of 96.5% pure CO, how many grams of excess reactant are left unreacted?
-
The atmosphere is composed of 21% molecular Oxygen and approximately 79% molecular Nitrogen. What is the molecular weight of dry air?
-
The atmosphere is composed of 21% molecular Oxygen and approximately 79% molecular Nitrogen. What is the molecular weight of dry air?
Using the molar mass of dry air calculated above, what is the mass fraction of Oxygen present in 10 moles of air?
-
AP edition reviewed standalone replacement.
Hypothetical model:1.000kg solution contains0.1000mol diprotic acidH₂A. Its first dissociation is complete and exactly10.0% ofHA⁻ undergoes the second dissociation. Find the formal amount ofH⁺ and formal hydrogen-ion mass percent usingM(H)=1.008g mol⁻¹. This is not a concentrated sulfuric-acid claim.
-
The decomposition of organic matter (CH2O) to produce methane can be simplified by the following unbalanced reaction:
1 kg of organic matter is added to an enclosed environment that has 10 moles of Oxygen present.
What is the mass percent of Carbon before and after the reaction? (Hint: Do not violate the law of conservation of mass) -
The decomposition of organic matter (CH2O) to produce methane can be simplified by the following unbalanced reaction:
1 kg of organic matter is added to an enclosed environment that has 10 moles of Oxygen present.
What is the mass percent of Carbon before and after the reaction? (Hint: Do not violate the law of conservation of mass)Based on the above reaction, what is the mass percent of methane (CH4) once the Oxygen is used up?
-
The decomposition of organic matter (CH2O) to produce methane can be simplified by the following unbalanced reaction:
1 kg of organic matter is added to an enclosed environment that has 10 moles of Oxygen present.
What is the mass percent of Carbon before and after the reaction? (Hint: Do not violate the law of conservation of mass)Based on the reaction in #14, what is the mass percent of CO2 in the container after the Oxygen has reacted?
-
AP edition reviewed replacement.
Reviewed ideal material-balance replacement: completely react1.000kg glucose with exactly stoichiometric oxygen from dry air modeled as21.0mol%O₂ and79.0mol%N₂. UseC₆H₁₂O₆+6O₂→6CO₂+6H₂O. Nitrogen passes unchanged and all products remain gaseous. This is a stated ideal bookkeeping model. Use molar masses glucose180.156,O₂32.00,N₂28.014,CO₂44.01,H₂O18.016g mol⁻¹ and dry-air density1.20gL⁻¹.
Find the volume of required dry air in cubic meters. Show how oxygen demand determines total air amount.
-
AP edition reviewed replacement.
Reviewed ideal material-balance replacement: completely react1.000kg glucose with exactly stoichiometric oxygen from dry air modeled as21.0mol%O₂ and79.0mol%N₂. UseC₆H₁₂O₆+6O₂→6CO₂+6H₂O. Nitrogen passes unchanged and all products remain gaseous. This is a stated ideal bookkeeping model. Use molar masses glucose180.156,O₂32.00,N₂28.014,CO₂44.01,H₂O18.016g mol⁻¹ and dry-air density1.20gL⁻¹.
Find the CO₂ mass percent in the complete gaseous product mixture, including unchanged nitrogen.
-
AP edition reviewed replacement.
Reviewed ideal material-balance replacement: completely react1.000kg glucose with exactly stoichiometric oxygen from dry air modeled as21.0mol%O₂ and79.0mol%N₂. UseC₆H₁₂O₆+6O₂→6CO₂+6H₂O. Nitrogen passes unchanged and all products remain gaseous. This is a stated ideal bookkeeping model. Use molar masses glucose180.156,O₂32.00,N₂28.014,CO₂44.01,H₂O18.016g mol⁻¹ and dry-air density1.20gL⁻¹.
Reviewed counting-entity revision: find the final mole fraction ofO₂ molecules. Explain why oxygen atoms can still be present when that molecular fraction is zero.
-
AP edition reviewed replacement.
Reviewed ideal material-balance replacement: completely react1.000kg glucose with exactly stoichiometric oxygen from dry air modeled as21.0mol%O₂ and79.0mol%N₂. UseC₆H₁₂O₆+6O₂→6CO₂+6H₂O. Nitrogen passes unchanged and all products remain gaseous. This is a stated ideal bookkeeping model. Use molar masses glucose180.156,O₂32.00,N₂28.014,CO₂44.01,H₂O18.016g mol⁻¹ and dry-air density1.20gL⁻¹.
Compare the oxygen-atom mass percent in incoming dry air with that in the product mixture. Report the change in percentage points and account for the added glucose oxygen.