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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:
See explanation
Explanation:
Atomic size increases down the group due to the addition of more shells.
As more shells are added and repulsion of inner electrons become more significant, atomic size increases down the group. However, across the period, atomic size decreases due to increase in effective nuclear charge without any increase in the number of shells. This causes increased attraction between the nucleus and the outermost shell thereby decreasing the size of the atom.
Ionization energy decreases down the group because the outermost electron is more shielded by inner electrons making it easier for this outermost electron to be lost. Across the period, ionization energy increases due to increase in effective nuclear charge which makes it more difficult to remove the outermost electron due to increased nuclear attraction.
So the empirical formula is Mg3N2
Explanation:
from the equation 1 mole of O2 will give 2 moles of H2O then 6.0 moles of O2 will give x
6.0*2 moles/ 1 mole
= 12 moles
this implies that, 6.0 moles of O2 will give = 12 moles of water
