Question #1199a

2 Answers
Andrea B.
Apr 20, 2017

0,9874 g of iron (III) oxide will be produced

Explanation:

in all these kind of problems you must find two thing: 1) the ratio of combination in the reaction between the two species that react ( water and iron (III) oxide). 2) the number of mol through the gas law (PV= nRT)
1) #2Fe +3 H_2O = Fe_2O3+ 3H_2#
2) #n= (PV)/(RT)= (0,121 atm 4 L)/ (0,082 ( L atm)/(mol K) (50,2 +273,2)K))= 0,0183 mol of #H_2O# that give 0,0061 mol of iron (III) oxide (ratio is 3:1) The weigt of 0,0061 mol of #Fe_2O3# is 0,0061mol x159,7 (g/mol) = 0,9874 g

Andrea B.
Apr 20, 2017

in all these kind of problems you must find two thing: 1) the ratio of combination in the reaction between the two species that react ( water and iron (III) oxide). 2) the number of mol through the gas law (PV= nRT)
1) #2Fe +3 H_2O = Fe_2O3+ 3H_2#
2) #n= (PV)/(RT)= (0,121 atm 4 L)/ (0,082 ( L atm)/(mol K) (50,2 +273,2)K)= 0,0183 mol# of #H_2O# that give 0,0061 mol of iron (III) oxide (ratio is 3:1)
The weigt of 0,0061 mol of #Fe_2O3# is 0,0061mol x159,7 (g/mol) = 0,9874 g

Related questions
See all questions in Gas Stoichiometry
Impact of this question
2740 views around the world
You can reuse this answer
Creative Commons License