QUIZ #9, CHEM 001  (Chapter 12)  

 Name:___________________________ Date:___________  

You must show all work to receive any partial credit and, where applicable, report all numerical answers  to the proper number of significant figures.  

1. How many grams of NH 3 are contained in 1.50 L of the gas at a temperature of 165°C  and a pressure of 250. torr? [8 pts.]  

 

 

 

2. What is the molar mass of an unknown gas if 0.948 g of the gas occupies 3.5 L at 65°C  and 250. torr? [9 pts.]  

   

 

 

 

3 .   How many liters of H 2 will be produced when 30.49 g of Na react with 27.03 g of  H 2 O?  After the reaction had gone to completion with an 85.0 % yield, the   temperature and  pressure were found to be 123 ° C and 533 torr, respectively.   [8 pts.]   

Na (s) + H 2 O (l) ⎯→ NaOH (aq) + H 2 (g)    

 

 

ANSWER KEY

PV =

n RT 250.

torr = 0.329 atm

n = PYRT 165°C = 438 K g17.04 glmol

n = ( 0.329 atm )( 1.504/10.0821 Loatmlmolok) ( 438K ) = 0.0137 mot NHS

0.0137 mot NHS x 17.04 glmal = 0.233 grams NHS

250 .torr = 0.329 atm

65°C = 338 K

n = PVIRT= ( 0.329 atm ) ( 3.54/10.0821 L.atmlmol.lt) ( 338k ) = 0.0415 mot

0.948g 10.0415 mot = 22.8 glmal

2 2 2

533 torn = 0.701 atm 123°C = 396 K2 Na

30.49g Na x

gigx

lmdHI

=O.6631m42ft / !¥

"

2 mot Nalimiting reactant

im×

I mot H2= 0.7500 mol He27.03g H2O ×

18.02g 2 mot H2O

( 0.6631 mot Hz) ( O .

850)=0.564mot Hz

pv = n RT

✓ = nRT/p = ( 0.5636 mot Hz) ( 0.0821 L . atmlmol . K) ( 396K ) / ( 0.701 atm )

V = 26.1 L Hz

QUIZ #9, CHEM 001  (Chapter 12)  

 Name:___________________________ Date:___________  

You must show all work to receive any partial credit and, where applicable, report all numerical answers  to the proper number of significant figures.  

1. How many grams of NH 3 are contained in 1.05 L of the gas at a temperature of 165°C  and a pressure of 250. torr? [8 pts.]  

 

 

 

2. What is the molar mass of an unknown gas if 0.748 g of the gas occupies 0.5 L at 65°C  and 250. torr? [9 pts.]  

   

 

 

 

3 .   How many liters of H 2 will be produced when 20.49 g of Na react with 13.03 g of  H 2 O?  After the reaction had gone to completion with an 83.0 % yield, the   temperature and  pressure were found to be 123 ° C and 533 torr, respectively.   [8 pts.]   

Na (s) + H 2 O (l) ⎯→ NaOH (aq) + H 2 (g)    

 

 

ANSWER HEY

PV =

n RT 250.

torr = 0.329 atm

n = PYRT 165°C = 438 K g17.04 glmol

n = ( 0.329 atm )(

1.054/10.0821

L .atmlmolok) ( 438K ) = 0.00961 mot NHS

0.00961 mot NHS x 17.04 glmal = 0.164 grams NHS

250 .torr = 0.329 atm

65°C = 338 K

n -

- PVIRT= ( 0.329 atm ) (

0.54/10.0821L.atmlmol.lt) ( 338k ) =

0.00593mot

0.748g10.00593mot = 126.14 glmal

2 2 2

533 torn = 0.701 atm 123°C = 396 K20.49gNa x xlmdH2_

= 0.446 mol Hz22.99g 2 mot Na13.03g

H2O ¥x

I mot H2=

O.36I5mo×

18.02g 2 mot H2Olimiting reactant

(

0.3615mot Hz) (

0.8301=0.300mot Hz

pv = NRT

✓ = nRT/p = ( 0.300 mot Hz) ( 0.0821 L.at/mol.k)(396K)/( 0.701 atm )

V = 13.9 L Hz

QUIZ #9, CHEM 001  (Chapter 12)  

 Name:___________________________ Date:___________  

You must show all work to receive any partial credit and, where applicable, report all numerical answers  to the proper number of significant figures.  

1. How many grams of NH 3 are contained in 0.55 L of the gas at a temperature of 165°C  and a pressure of 250. torr? [8 pts.]  

 

 

 

2. What is the molar mass of an unknown gas if 0.908 g of the gas occupies 2.5 L at 65°C  and 250. torr? [9 pts.]  

   

 

 

 

3 .   How many liters of H 2 will be produced when 30.49 g of Na react with 15.03 g of  H 2 O?  After the reaction had gone to completion with an 92.0 % yield, the   temperature and  pressure were found to be 123 ° C and 533 torr, respectively.   [8 pts.]   

Na (s) + H 2 O (l) ⎯→ NaOH (aq) + H 2 (g)    

 

 

ANSWER KEY

PV =

n RT 250.

torr = 0.329 atm

n -

- PYRT 165°C = 438 K g17.04 glmol

n = ( 0.329 atm )(

0.554/10.0821

Loatmlmolok) ( 438K ) = 0.00503 mot NHS

0.00503 mot NHS x 17.04 glmal = 0.0857 grams NHS

250 .torr = 0.329 atm

65°C = 338 K

n -

- PVIRT= ( 0.329 atm ) (

2.54/10.0821L.atmlmol.lt) ( 338k ) =

0.0296mot

0.908g10.0296mot = 30.7 glmal

2 2 2

533 torn = 0.701 atm 123°C = 396 K30.49gNa ×

zgj×lzmom!a= 0.663 mot Hz15.03g

H2O ¥×

I mot H2=

O.4l7mo×

18.02g 2 mot H2Olimiting reactant

(

0.417mol Ha) (

0.9201=0.384mot Hz

pv = NRT

✓ = nRT/p = (

0.384mot Hz) ( 0.0821 L . atmlmol . K) ( 396K ) / ( 0.701 atm )

✓ = 17.8 L Hz