Answer:- Third choice is correct, 17.6 moles
Solution:- The given balanced equation is:
Al_2(SO_4)_3+6KOH\rightarrow 2Al(OH)_3+3K_2SO_4
We are asked to calculate the moles of potassium hydroxide needed to completely react with 2.94 moles of aluminium sulfate.
From the balanced equation, there is 1:6 mol ratio between aluminium sulfate and potassium hydroxide.
It is a simple mole to mole conversion problem. We solve it using dimensional set up as:
2.94molAl_2(SO_4)_3(\frac{6molKOH}{1molAl_2(SO_4)_3})
= 17.6 mol KOH
So, Third choice is correct, 17.6 moles of potassium hydroxide are required to react with 2.94 moles of aluminium sulfate.
An atom or molecule with a net electric charge due to the loss or gain of one or more electrons. :)
Given:
Concentration of titrant = 0.1000 M
Volume of titrant = 45 mL
The molarity of analyte depends on the amount of the analyte present in the titrated solution. If the amount of analyte is 20 mL, then its concentration is:
45ml * 0.10 M = C analyte * 20 ml
C analyte = 0.225 M
1 mols of Aluminium ion forms 1 mol aluminium phosphate
Molar mass of AlPO_4
Moles of AlPO_4
- 61µg/106
- 0.000061/106
- 5.75×10^{-7}
- 57.5µmol
Moles of Al3+=57.5µmol
Answer:
what I got was 0.8435160945347224 moles
