# Calculate the mass of iron in the original mixture!

I seemingly can't solve the following problem, I've spend over 30 mins on it, no progress, can any of you geniusses give a step by step solution to me? Thank you so much

Problem:

A mixture of aluminium and iron weighing 9,62g reacts with hydrogen chloride in aqueos solution according to the parallel reactions

2Al + 6HCL --> 2AlCl3 + 3H2 Fe + 2HCL --> FeCl2 + H2

A 0,0738 quantity of hydroen is evolved when the metals react completely. Calculate the mass of iron in the original mixture.

Here's the solution I found on Slader.com

1) x + y = 9.62g; Considering that x is a mass of Al, and y is the mass of Fe

2) ; Ar (Al) = 26.98 g/mol ;Ar (Fe) = 55.85 g/mol

5) 0.0556x + 0.0377y = 0.366

x + y = 9.62 (multiplying with 0.0566)

0.0566x + 0.0566y = 0.535

Substracting this two equations

0.0377y = 0.169

y = 4.48 g Fe

6) x + 4.48 = 9.62

x = 5.14 g Al

-

Can someone explain step 5 to me? I really don't understand how one gets the "0.0377y = 0.169".

Thanks so much!!

Well, I understand it, but I don't like it. There's an error in line 5 of eq 5.

When you subtract the two equations, put everything in writing:

0.0556x + 0.0377y - 0.0566x -0.0566y = 0.366 - 0.535

Then -0.0189y = -0.169, (not 0.0377y = 0.169)

Then y = 8.94g (Fe) (answer)

(And incidentally, x + y = 9.62g, so x = 0.68g (Al))

I sympathize with you - but taking a little shortcut sometimes makes it difficult to detect an error. Sometimes you just have to write out everything.