Question

How many grams of Al2(Cr)4)3 can be produced from 125 mL of a 0.150 Molarity K2CrO4 solution?
Balanced equation: 2 Al(NO3)3(aq) + 3K2CrO4(aq) = Al2(CrO4)3(s) + 6KNO3(aq)
Molar Mass: 213 194 402 111

Answer 1

Ok so do you know where to start?

Answer 2

We first have to find the number of moles of \(K_2CrO_4\) solution, then use that to switch using mol to mol conversion.

Answer 3

ok so .125L * 0.150 M = .01875 moles K2CrO4

Answer 4

Now you can convert that using mol to mol conversion using the coefficients on the reaction equation,
\[0.01875mol~K_2CrO_4 \times \frac{1~mol~Al_2(CrO_4)_3}{3~mol~K_2CrO_4 }=\]

Answer 5

After that, you'll have \(~moles~of~Al_2(CrO_4)_3\). Then multiply it by the relative molecular mass to get the mass of \(Al_2(CrO_4)_3\)

Answer 6

okay awesome!! i got it now! thanks.. it can get confusing when i can't find an example in my notes close enough to figure out all the steps to go by

Answer 7

Oh you're glad you've got one here :)