Chem,1411 Study Guide Exam 2 2018.Spring - Protected View- Saved to this PC ence

Question

Chem,1411 Study Guide Exam 2 2018.Spring - Protected View- Saved to this PC ences Mailings Review View Help Tell me what you want to do in viruses. Unless you need to edit, it's safer to stay in Protected View. Enable Editing is plenty of water present? 3 NO2(g) + H20(1) 2 HNO3(aq) + NO(g) A) 15.72 moles NO2 B) 25.3 moles NO C) 8.44 moles NO D) 5.63 moles NO2E) 1.83 moles NO 2. How many grams of NO are formed when 6.21 moles of NO; are reacted? A) 19.9 g NO B) 279gNO C) 62.1 g NO D)3.59 g NO E1.7 g NO 3. If 14.01 g sample of N2 reacts with 4.02 g of H2 to forn ammonia (NH3), what is the mass of les NO: E) 1.83 moles NO 3 NO2(B)+H20(0) 2 HNO3(ag)+NO(B) ammonia formed? A) 34.02g B) 11.10 g C) 17.01g D) 30.02 E) 23.07 g 4. The chemical equation shows that carbonic acid is decomposed to form water and carbon dioxide by heating. What is the mass of carbon dioxide formed from 12.40 g of carbonic acid? A) 7.58 g B) 8.80 g C) 2.20 g D) 6.20 5. How many milliliters of a 0.294 M LiOH solution contain 0.223 moles ofLiOH? A) 543 mL. B) 163 ml c) 759 ml D) 885 ml. )326 ml.

Explanation / Answer

there are 11 questions ...

I am solving first 3 questions ....remaining questions should be re-posted separately individually.

we will be happy to solve the remaining questions also.

1) as per the balanced reaction,

for each mole of NO , we need 3 moles of NO2

so for 5.24 moles of NO, we need 3 X 5.24 = 15.72 moles of NO2

Option A

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2) as per balanced reaction.

3 moles of NO2 , produces/forms 1 mole of NO

so for 6.21 moles of NO2, we need 1/3( moles of NO)= 1/3( 6.21) moles =2.07 moles

mass of 2.07 moles of NO = 2.07 x molar mass of NO = 2.07 x 30 = 62.1 grams ANS

OPTION C

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3) Balanced reaction for formation of NH3 ( ammonia )

N2+ 3H2 -------> 2NH3

moles of Nitrigen = 14.02/28 = 0.5 moles

moles of Hydrogen = 4.02/2 = 2.01 moles

nitrogen is in limited quantity....

so its limiting reagnet or Deciding factor for product amount as it will be used up first.

moles of Ammonia = 2 x moles of nitrogen = 0.5 x 2 = 1 moles = 17.01 grams of nitrogen.

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