What volume of h2o is formed at stp when 6.0g of al is treated with excess naoh? naoh + al + h2o —-> naal(oh)4 + h2 (g)

Answer : The volume of is 14.784 L.

Solution : Given,

Mass of Aluminium = 6 g

Molar mass of Aluminium = 27 g/mole

First we have to calculate the moles of aluminium.

Moles of Al =

The given balanced reaction is,

From the reaction, we conclude that

2 moles of Al react with the 6 moles of

0.22 moles of Al react with of

At STP,  1 mole contains 22.4 L volume

As, 1 mole of contains 22.4 L volume of

0.66 moles of contains volume of

Therefore, the volume of is 14.784 L.                                                                                        

the answer would be a