Question

Question #1199a

Answer 1

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

Answer 2

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