739 is the number of atoms of O in 92.3 moles of Cr3(PO4)2.
Explanation:
Molecular formula given is = Cr3(PO4)2
number of moles of the compound is 92.3 moles
number of 0 atoms in 92.3 moles =?
From the chemical formula 1 mole of the compound has 8 atoms of oxygen
So, it can be written as
1 mole Cr3(PO4)2 has 8 atoms of oxygen
92.3 moles of Cr3(PO4)2 has x atoms of oxygen
=
x = 8 x 92.3
x = 738.4 atoms
There will be 739 oxygen atoms in the 92.3 moles of Cr3(PO4)2.
Answer : 51.8 g of nitrogen are needed to produce 100 grams of ammonia gas.
Solution : Given,
Mass of = 100 g
Molar mass of = 27 g/mole
Molar mass of = 28 g/mole
First we have to calculate moles of .
The given balanced chemical reaction is,
From the given reaction, we conclude that
2 moles of produced from 1 mole of
3.7 moles of produced from
of
Now we have to calculate the mass of .
Mass of = Moles of
× Molar mass of
Mass of = 1.85 mole × 28 g/mole = 51.8 g
Therefore, 51.8 g of nitrogen are needed to produce 100 grams of ammonia gas.