To 100 mL of 5 M NaOH solution (density 1.2 g/mL) were added 200 mL of another NaOH solution which has a density of 1.5 g/mL and contains 20 mass percent of NaOH. What will be the volume of the gas (at STP) in litres liberated when aluminium reacts with this (final) solution. The reaction is Al + NaOH + H2O → NaAlO2 + 3/2 H2 (At. wt. Na = 23, O = 16, H = 1)

Question

To 100 mL of 5 M NaOH solution (density 1.2 g/mL) were added 200 mL of another NaOH solution which has a density of 1.5 g/mL and contains 20 mass percent of NaOH. What will be the volume of the gas (at STP) in litres liberated when aluminium reacts with this (final) solution. The reaction is Al + NaOH + H2O → NaAlO2 + 3/2 H2 (At. wt. Na = 23, O = 16, H = 1) (A) 67.2 L (B) 89.6 L (C) 44.8 L (D) 22.4 L

Tardigrade Admin · Accepted Answer

Moles of NaOH in 1st solution = 0.5 mol Moles of NaOH added =(200×1.5×0.2/40)=1.5 Moles of NaOH in the final solution = 1.5 + 0.5 = 2 mol Al+NaOH+H2O arrow AaAlO2+(3/2)H2 Moles of H2 produced from 2 mol of NaOH = 3 mol Volume of H2 produced at STP = 3 Ã 22.4 = 67.2 L

67.2 L

89.6 L

44.8 L

22.4 L

Moles of NaOH in 1st solution =0.5mol Moles of NaOH added =40200×1.5×0.2​=1.5 Moles of NaOH in the final solution =1.5+0.5=2mol Al+NaOH+H2​O→AaAlO2​+23​H2​ Moles of H2​ produced from 2mol of NaOH=3mol Volume of H2​ produced at STP=3×22.4=67.2L