Question

Dont remember how to do this-

Na3PO4+3KOH=3NaOH+1K3PO4
If excess Na3PO4 reacts with 5g of KOH, how many grams of K3PO4 will be produced?

Answer 1

First you have to find the number of moles of KOH, then by using that you can find the number of moles of K3PO4 by multiplying coefficients

Answer 2

What??? O.o now im even more confused. I dont remeber having to do that. All i can really remember is that you cross out "if excess Na3PO4" and go from there

Answer 3

You basically don't have to care about other reactants except for KOH and K3PO4, because these 2 are the one that matters, KOH is excess so we don't need to find which is limiting.

The trick here is to use the moles of KOH to find the moles of K3PO4
\[5gKOH \times\frac{1molKOH}{56gKOH} \times[\frac{1molK_3PO_4}{3molKOH}] \times \frac{212gK_3PO_4}{1molK_3PO_4}=6.31gK_3PO_4\]
The middle part of the calculation is the mole conversion, can you see it now?

Answer 4

Yes thank you