Question

The metathesis reaction between sodium hydroxide and aluminum nitrate is shown below

3NaOH(aq)+ Al(NO3)3(aq) arrow Al(OH)3(s)+ 3 NaNo3(aq)

How many grams of solid aluminum hydroxide can be produced by the reaction of 375.0ml of 0.400M sodium hydroxide solution with 250.0ml of 0.150M aluminum nitrate nitrate solution?

Answer 1

volume OF NaOH, V = 375.0 mL
= 0.375 L


number of mol OF NaOH,
n = Molarity * Volume
= 0.4*0.375
= 0.15 mol
volume OF Al(NO3)3 , V = 250.0 mL
= 0.25 L


number of mol OF Al(NO3)3,
n = Molarity * Volume
= 0.15*0.25
= 3.75*10^-2 mol
Balanced chemical equation is:
3 NaOH + Al(NO3)3 ---> Al(OH)3 + 3 NaNO3


3 mol of NaOH reacts with 1 mol of Al(NO3)3
for 0.15 mol of NaOH, 0.05 mol of Al(NO3)3 is required
But we have 0.0375 mol of Al(NO3)3

so, Al(NO3)3 is limiting reagent
we will use Al(NO3)3 in further calculation


Molar mass of Al(OH)3,
MM = 1*MM(Al) + 3*MM(O) + 3*MM(H)
= 1*26.98 + 3*16.0 + 3*1.008
= 78.004 g/mol

According to balanced equation
mol of Al(OH)3 formed = (1/1)* moles of Al(NO3)3
= (1/1)*0.0375
= 0.0375 mol


mass of Al(OH)3 = number of mol * molar mass
= 3.75*10^-2*78
= 2.925 g
Answer: 2.925 g