How many moles of aluminum are required to completely react with 107 mL of 6.00 M H.SO. according to the balanced chemical reaction: 2 Al(s) + 3 H:SO.(aq) → Al:(SO.)»(aq) + 3 H:(g)

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Chapter5: Gases
Section: Chapter Questions
Problem 161IP: In the presence of nitric acid, UO2+ undergoes a redox process. It is converted to UO22+ and nitric...
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How many moles of aluminum are required to completely react with 107
mL of
6.00 M H2SO. according to the balanced chemical reaction:
2
Al(s) +
3
H:SO:(aq) → Al:(SO.)>(aq) +
H:(g)
X
STARTING AMOUNT
ADD FACTOR
ANSWER
RESET
*( )
0.428
0.0374
1
107
1000
0.001
0.963
98.08
6.022 x 1023
428
6.00
mL H2SO4
M H2SO4
g H2SO4
g AI
mol H2SO. g A:(SO4)3
L H2SO4
mol H2
mol Al
mol Al:(SO4)s
g H2
2.
Transcribed Image Text:How many moles of aluminum are required to completely react with 107 mL of 6.00 M H2SO. according to the balanced chemical reaction: 2 Al(s) + 3 H:SO:(aq) → Al:(SO.)>(aq) + H:(g) X STARTING AMOUNT ADD FACTOR ANSWER RESET *( ) 0.428 0.0374 1 107 1000 0.001 0.963 98.08 6.022 x 1023 428 6.00 mL H2SO4 M H2SO4 g H2SO4 g AI mol H2SO. g A:(SO4)3 L H2SO4 mol H2 mol Al mol Al:(SO4)s g H2 2.
Question 22 of 27
How many moles of aluminum are required to completely react with 107
mL of
6.00
M H2SO4 according to the balanced chemical reaction:
Al(s) +
3
H:SO:(aq) – Al:(SO.)>(aq) +
H:(g)
STARTING AMOUNT
ADD FACTOR
ANSWER
RESET
*( )
0.428
0.0374
1
3
107
1000
0.001
0.963
98.08
6.022 x 1023
428
6.00
mL H2SO4
M H2SO4
g H2SO4
g AI
mol H2SO4 g Al:(SO4)3
L H2SO4
mol H2
mol Al
mol Al:(SO.)s
g H2
Transcribed Image Text:Question 22 of 27 How many moles of aluminum are required to completely react with 107 mL of 6.00 M H2SO4 according to the balanced chemical reaction: Al(s) + 3 H:SO:(aq) – Al:(SO.)>(aq) + H:(g) STARTING AMOUNT ADD FACTOR ANSWER RESET *( ) 0.428 0.0374 1 3 107 1000 0.001 0.963 98.08 6.022 x 1023 428 6.00 mL H2SO4 M H2SO4 g H2SO4 g AI mol H2SO4 g Al:(SO4)3 L H2SO4 mol H2 mol Al mol Al:(SO.)s g H2
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