Question

What volume of oxygen gas at 35.0°C and 1.03 atm is required to react completely with 17.0 g of iron according to the following reaction? iron(s) + oxygen(g) iron(II) oxide(s)

___ liters oxygen gas

Answer 1

Molar mass of Fe = 55.85 g/mol

mass of Fe = 17.0 g

we have below equation to be used:

number of mol of Fe,

n = mass of Fe/molar mass of Fe

=(17.0 g)/(55.85 g/mol)

= 0.3044 mol

the reaction taking place is:

2 Fe + O2 —> 2 FeO

from reaction,

moles of O2 required = (1/2)*moles of Fe

= (1/2)*0.3044 mol

= 0.1522 mol

we have:

P = 1.03 atm

n = 0.1522 mol

T = 35.0 oC

= (35.0+273) K

= 308 K

we have below equation to be used:

P * V = n*R*T

1.03 atm * V = 0.1522 mol* 0.0821 atm.L/mol.K * 308 K

V = 3.74 L

Answer: 3.74 L

Feel free to comment below if you have any doubts or if this answer do not work. I will correct it and submit again if you let me know