H3PO4(aq) + 3KOH(aq) -> K3PO4(aq) + 3H2O(l) - molarity?

Question

What is the concentration of the original KOH solution if 84.0 mL of the KOH solution reacts with 3.40 g of H3PO4?

Answer 1

Lets see...
Molecular weight of H3PO4: 30.978 g/mol
Divide this to 3.4g (meaning 3.4/30.978) to get: 0.109755 mol
Now you have the molarity of H3PO4.
Then, we balance the reaction: already balanced
Next, we assume that the reaction is 100% complete, so the moles of H3PO4=3x(moles of KOH). Value is: 0.32926593
Convert first KOH volume to Liters: 84/1000
then divide this to moles of KOH to get:
3.9198M-> final answer
It is beneficial to do cancelation of units when answering stoichiometric questions.

Answer 2

This is a stoichiometry question.

Convert the 3.40 g to moles of H3PO4

Multiply by the molar ratio to get moles of KOH. This gives you moles of KOH

1 molar = 1 mole / 1000 ml of solution.

Once you have the moles, you can calculate the molarity of 84.0 ml. Just use a ratio