M1V1 = M2V2
M1 = 3.000 M
V1 = 0.8000 L
M2 = ?
V2 = 2.00 L
M2 = M1V1/V2 = (3.000 M)(0.8000 L)/(2.00 L) = 1.20 M
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
The answers are:
1. D
2. B if it is a check all that are true it is b & d
Answer:
Explanation:
n CaCO3 = mass / m.wt
= 500 /( 40 + 12 + 16x 3)
= 5 mole
n CaO = 5 moles ( from the balanced equation we have 1:1 moles )
mass of CaO = nCaO X m.wt
5 x( 40 +16 )
= 280 grams
