150 cm3 0.24 M naoh solution react with excess al(no3)3, calculate the mass of saodium aluminate produced.
al(no3)3 + 4naoh -> naalo2 + 3 nano3 +2h20
i got 0.738g..
is my answer correct? if im wrong could u show me the calculation?
Please use conventional symbols as this makes things easier to read and you also practise with the 'language' of the subject.
Al(NO3)3 + 4NaOH -> NaAlO2 + 3 NaNO3 +2H2O
So 4 moles of NaOH produce 1 moles of NaAlO2.
or
So 1 mole of NaOH produce 1/4 moles of NaAlO2.
We start with
0.150 L x 0.24 mole L^-1 = 0.036 mole
this must produce
0.036 mole/4 moles of NaAlO2
=0.009 moles
The relative molecular mass of NaAlO2 is
23+27+32 = 82
so the mass produced is
0.009 moles x 82 g mole^-1
=0.738 g
so you are correct.