Question

6 NaOH + 2 Al ???? 2 Na3AlO3 + 3 H2 How much aluminum is required to produce 17.5 grams of hydrogen? How many moles of NaOH are required to produce 3.0 grams of hydrogen? How many moles of hydrogen can be prepared from 1.0 grams of aluminum?

Answer 1

Answer:

a) 157.5 grams of aluminum.

b) 1 mol

c) 9 g

Explanation:

The reaction is :

$6 NaOH + 2Al ---> 2 Na_{3}AlO_{3} + 3H_{2}$

As per balanced equation

a) 3 moles of hydrogen will be produced from two moles of aluminium.

The atomic mass of aluminium = 27

therefore

3X2 grams of hydrogen is produced from 2 X 27 grams of Al

1 gram of hydrogen will be produced from $\frac{2X27}{3X2}= 9$g

therefore 17.5 will be produced from = 9X 17.5 = 157.5 grams of aluminum.

b) as per balanced equation three moles or six gram of hydrogen is produced from 6 moles of NaOH.

Therefore 1 g of hydrogen will be produced from =$\frac{6}{6}$

or 1 gram will be prepared from = 1 mole

c) from balanced equation three moles are produced from two moles of Al (27X2 = 54 g).

thus from 54  grams gives 6 grams of hydrogen

1 grams will give = $\frac{54X1}{6}= 9 g$

Answer 2

Answer:

157.5 g Al is required to produce 17.5 g of hydrogen

3 moles of NaOH are required to produce 3 g of hydrogen

0.055 moles of Hydrogen can be prepared from 1 g of Al.

Explanation:

The equation of reaction is:

6 NaOH + 2 Al    ----->   2 Na3AlO3 + 3H2

For mass of Aluminum:

It is clear from eqn. that:

3 (Molar Mass of H2) requires = 2 (Atomic Mass of Al)

3(2 g H2) requires = 2 (27 g) Al

1 g H2 requires = (54/6) g Al

17.5 g H2 requires = (17.5 x 9) g Al

17.5 g H2 requires = 157.5 g Al

For moles of NaOH:

It is clear from eqn. that:

3 (Molar Mass of H2) requires = 6 moles of NaOH

3(2 g H2) requires = 6 moles of NaOH

1 g H2 requires = (6/6) moles of NaOH

3 g H2 requires = 3 x 1 moles of NaOH

3 g H2 requires = 3 moles of NaOH

For moles of H2:

It is clear from eqn. that:

2 (Molar Mass of Al) produce = 3 moles of H2

2(27 g Al) produce = 3 moles of H2

1 g Al produce = (3/54) moles of H2

1 g Al produce = 0.055 moles of H2