Answer:
We'll have 1 mol Al2O3 and 3 moles H2
Explanation:
Step 1: data given
Numer of moles of aluminium = 2 moles
Number of moles of H2O = 6 moles
Step 2: The balanced equation
2Al + 3H2O → Al2O3 + 3H2
Step 3: Calculate the limiting reactant
For 2 moles Al we need 3 moles H2O to produce 1 mol Al2O3 and 3 moles H2
Aluminium is the limiting reactant. It will completely be consumed (2 moles).
H2O is in excess. There will react 3/2 * 2 = 3 moles
There will remain 6 - 3 = 3 moles
Step 4: Calculate moles products
For 2 moles Al we need 3 moles H2O to produce 1 mol Al2O3 and 3 moles H2
For 2 moles Al we'll have 2/1 = 1 mol Al2O3
For 2 moles Al We'll have 3/2 * 2 = 3 moles H2
We'll have 1 mol Al2O3 and 3 moles H2
Answer:
The answer to your question is letter B. 9
Explanation:
Unbalanced reaction
Al₂(SO₄)₃ + Ca(OH)₂ ⇒ Al(OH)₃ + CaSO₄
Reactants Elements Products
2 Al 1
3 S 1
14 O 7
1 Ca 1
2 H 3
Balanced reaction
Al₂(SO₄)₃ + 3Ca(OH)₂ ⇒ 2Al(OH)₃ + 3CaSO₄
Reactants Elements Products
2 Al 2
3 S 3
18 O 18
3 Ca 3
6 H 6
The sum of the coefficients is 1 + 3+ 2+ 3 = 9
