Find the mass of iron used in the experiment.

[s3a]

Homework Statement

"A student reacts an unknown mass of iron (Fe) in an excess of sulfuric acid
(H2SO4). The balanced equation is:
2 Fe(s) + 3sH2SO4(aq) → Fe2(SO4)3(aq) + 3 H2(g)
She records the following data:
23.0°C
Temperature
Pressure 98.4 kPa
Volume of dry hydrogen collected 47.3 mL
Find the mass of iron used in the experiment."

Homework Equations

PV = nRT and a ratio.

The Attempt at a Solution

The solution says (the formatting was ruined when I pasted this and I tried to fix it so if anything is left ruined by accident, sorry):
" Given: T = 23.0°C + 273 = 296 K; V = 0.0473 L; P = 98.4 kPa;
R = 8.31 kPa⋅L/mol⋅K, we can calculate the number of moles of hydrogen gas
produced by the ideal gas equation: n = PV/RT
n = (98.4 kPa)(0.0473 L)/(8.31 kPa⋅L/mol⋅K)(296 K)
n = 1.89 x 10^(-3) mol of H2
From the balanced equation:
2 Fe(s) + 3H2SO4(aq) → Fe2(SO4)3(aq) + 3 H2(g) hydrogen and iron
are in a ratio of 3:2 moles. There must, therefore, be 2/3(1.91 x 10^(-3) mol) of Fe
present, that is, 1.26 x 10^(-3) mol of Fe. Since the molar mass of Fe is 55.85g/mol,
this represents 0.0705 g of Fe."

but I don't see how 1.89 x 10^(-3) mol becomes 1.91 x 10^(-3) mol nor do I see why the ratio is 2/3 of approximately the number of moles of H_2; I think it should be 3/2 of exactly the number of moles of H_2 as shown in my work.

My attempt is attached. I'm feeling that I'm correct but could someone please confirm if I am right and the solution is wrong or tell me why I am wrong if I am wrong?

Any input would be greatly appreciated!
Thanks in advance!

 

[chemisttree]

You are asked how many moles of IRON, not hydrogen. You know that the ratio of Fe2 is 2:3 and you know the exact value of hydrogen.

2Fe:3H2 or 2/3 the amount of hydrogen. ie. 2/3Fe:1H2

 

[s3a]

Isn't that what I did?:

n_Fe = 3/2 * n_(H_2)

(as shown in my attached work.)

 

[Saitama]

Seems like that its a mistake from where you have copied the solution. If you calculate the 2/3 of 1.89, you get 1.26.

 

[s3a]

Oh, I got it now. Thanks guys!