Respuesta :
Answer:
a) 157.5 grams of aluminum.
b) 1 mol
c) 9 g
Explanation:
The reaction is :
[tex]6 NaOH + 2Al ---> 2 Na_{3}AlO_{3} + 3H_{2}[/tex]
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 [tex]\frac{2X27}{3X2}= 9[/tex]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 =[tex]\frac{6}{6}[/tex]
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 = [tex]\frac{54X1}{6}= 9 g[/tex]
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
