Answer:
0.444 mol/L
Explanation:
First step is to find the number of moles of oxalic acid.
n(oxalic acid) = 
Now use the molar ratio to find how many moles of NaOH would be required to neutralize
of oxalic acid.
n(oxalic acid): n(potassium hydroxide)
1 : 2 (we get this from the balanced equation)
: x
x = 0.0111 mol
Now to calculate what concentration of KOH that would be in 25 mL of water:

Answer:
13.5 g
Explanation:
This question is solved easily if we remember that the number of moles is obtained by dividing the mass into the atomic weight or molar mass depending if we are referring to elements or molecules.
Therefore, the mass of aluminum in the reaction will the 0.050 mol Al times the atomic weight of aluminum.
number of moles = n = mass of Al / Atomic Weight Al
⇒ mass Al = n x Atomic Weight Al = 0.050 mol x 27 g mol⁻¹
= 13.5 g
We have three significant figures in 0.050 and therefore we should have three significant figures in our answer.
Beta particle in the reaction