# Question 1199a

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) $2 F e + 3 {H}_{2} O = F {e}_{2} O 3 + 3 {H}_{2}$
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

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) $2 F e + 3 {H}_{2} O = F {e}_{2} O 3 + 3 {H}_{2}$
2) $n = \frac{P V}{R T} = \frac{0 , 121 a t m 4 L}{0 , 082 \frac{L a t m}{m o l K} \left(50 , 2 + 273 , 2\right) K} = 0 , 0183 m o l$ of ${H}_{2} O$ that give 0,0061 mol of iron (III) oxide (ratio is 3:1)
The weigt of 0,0061 mol of $F {e}_{2} O 3$ is 0,0061mol x159,7 (g/mol) = 0,9874 g