6naoh + 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?

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