Question

what mass of al(oh)3 is produced if 22.7 g of naoh is consumed

Answer 1

The produced mass of $Al(OH)_{3}$ is 14.76 gram

Given - Mass of NaOH

Find - Mass of $Al(OH)_{3}$

Solution - Firstly writing the balanced chemical reaction -

$Al_{2}(SO_{4})_{3}$ + 6NaOH --> 3$Na_{2}(SO_{4})$ + 2$Al(OH)_{3}$

So, available moles of NaOH we have -

Number of moles = mass/molar mass

Molar mass of NaOH = 40 gram of NaOH

Number of moles = 22.7/40

Number of moles = 0.5675

In the reaction, 6 moles of NaOH are consumed that result in production of 2 moles of $Al(OH)_{3}$

So, 6 moles of NaOH = 2 moles of $Al(OH)_{3}$

0.5675 moles of NaOH = 2/6*0.5675

Moles of $Al(OH)_{3}$ produced = 0.189 moles

Molar mass of $Al(OH)_{3}$ 78 gram/mole.

Now, finding the mass of $Al(OH)_{3}$ produced = 0.189 × 78

Mass of $Al(OH)_{3}$ produced = 14.76 gram

Thus, mass of $Al(OH)_{3}$ produced is 14.76 gram.

#spj4