EXAMPLES WORKSHEET#12

(Chapter 12)   

Name: ____________________________   Date: ___________________  

1. Convert 631 torr to atm. (Answer: 0.830 atm)    

 

    

2. What volume will 1.80 𝗑 10 24 molecules of SO 3 occupy at STP? (Answer: 67.0 L)  

 

 

 

 

   

3. How many moles of CO 2 are contained in 9.55 L at 45℃ and 752 torr? (Answer: 0.362 mol)            

 

 

 

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ANSWER KEY

631 torn x I atm= 0.830 atm

760 torr

p -- I atm

T = 273K

PV = NRT

V = NRI V= ( 2.99 med )( 0.0821 L.atmlmol.lt) ( 273K )P

( t atm )1. 80×1024 molecules x I mal

6. 022×1023 molecules V = 67.0 L 503

PV = NRT n =PI

n =

( 0.989 atm )( 9.55L )RT

( 0.0821 Loatmlmolok) ( 318K )752 torrxtffaothforr = 0.989 atm

n - 0.362 mal

45°C = 318K

4. In the lab, students generated and collected hydrogen gas according to the following  

equation:  

 Mg + 2HCl ⎯→ MgCl 2   + H 2  

 

a. How many L of hydrogen gas at T = 273K and P = 1.00 atm were generated from 42.9 g  of magnesium metal reacting with 5.00 g of HCl? (Answer: 39.4 L)                

b. If 0.825 mL of hydrogen gas at T = 273K and P = 1.00 atm were needed, how many  moles of HCl would be required? (Answer: 0.0736 mol)  

  

  

Page 2 of 2  

g35.45+1.01 = 36.46 glmal

-152.0g42.9g My ×

12mg!zmzggx

tmdH2_=1.76ma

I mol My limiting reactant

152.0gHCl x lmotH#

×1molH2_ 2.08 mal Hz

36.46g2 mot HCl

=

PV = NRT

V = nRT/P= ( 1.76 mal Hz ) ( 0.0821 L.atmfmol.lt) ( 273K ) / ( 1 atm )

✓ = 39.4 L

•

PV =n RT

n = PVI RT = ( Latin ) ( 0.825L ) / ( 0.0821L . atnymolok) ( 273K )n = O

.

0368mot Hz

0.0368mot Hz ×

2.LA#z-- 0.0736 mot HCl

WORKSHEET#12 (Chapter 12)  

 Name: ____________________________   Date: ___________________  

1. Convert 503 torr to atm. (Answer:)    

 

    

2. What volume will 3.20 𝗑 10 24 molecules of NH 4 occupy at STP? (Answer:)  

 

 

 

 

   

3. How many moles of CO 2 are contained in 4.05 L at 25℃ and 702 torr? (Answer:)            

 

 

 

Page 1 of 2  

ANSWER HE

503 torn x I atm=

. 662 atm760 torr

p -- I atm

T = 273K

PV = NRT

V = NRI V= ( 5.31 med )( 0.0821 t.am/moloK)( 273K )P

( Latin )3.20×1024molecules x I mal

6. 022×1023 molecules V = 119 L

PV = NRT n =PI

n=( 0.923 atm )(

4.05L)

RT( 0.0821 Loatmlmolok) (

298K)

702torn x1atm__ =

0.923 atm760 torr

n - O.

15 mal

25°C=

298k

4. In the lab, students generated and collected H 2 S gas according to the following equation:  

 

2 HCl (aq) + Na 2 S (aq) → H 2 S (gas) + 2 NaCl (aq)  

 

a. How many L of H 2 S gas at T = 201K and P = 5.00 atm were generated from 12.9 g of  Na 2 S metal reacting with 24.3 g of HCl? (Answer:)                

b. If 0.125 mL of H 2 S at T = 273K and P = 1.23 atm were needed, how many moles of HCl  would be required? (Answer:)  

  

  

Page 2 of 2  

g35.45+1.01 = 36.46 glmal

ragnar.

' It.

on:{ x

' in:'

.in?.s=:÷÷Y'24.3g HCl x lmdH#

× lmdH#= O

.

333 mal H2S

36.46g 2 mot HCl

PV = n RT

V = nRT/P= ( 0.165 med) ( 0.0821 L.atmfmol.lt ) ( 201K )/( 5.00 atm)

V = 0.545 L

•

✓ = 0.125 L

pv = n RT

n = PVIRT

n = ( 1.23 atm ) ( 0.125 L )/( 0.0821 L.atmlmol.lt ) ( 273K )

n = 0.00686 mal H2S

0.00686 mal H2S × 2mdH# = O .0137 mot HCl

I not H2S