•3.9g of ammonia
•molar mass of ammonia = 17.03g/mol
1st you have to covert grams to moles by dividing the mass of ammonia with the molar mass:
(3.9 g)/ (17.03g/mol) = 0.22900763mols
Then convert the moles to molecules by multiplying it with Avogadro’s number:
Avogadro’s number: 6.022 x 10^23
0.22900763mols x (6.022 x 10^23 molecs/mol)
= 1.38 x 10^23 molecules
Answer:
4.43 g of Oxygen
Explanation:
As shown in Chemical Formula, one mole of Aluminium Sulfate [Al₂(SO₄)₃] contains;
2 Moles of Aluminium
3 Moles of Sulfur
12 Moles of Oxygen
Also, the Molar Mass of Aluminium Sulfate is 342.15 g/mol. It means,
342.15 g ( 1 mole) of Al₂(SO₄)₃ contains = 192 g (12 mole) of O
So,
7.9 g of Al₂(SO₄)₃ will contain = X g of O
Solving for X,
X = (7.9 g × 192 g) ÷ 342.15 g
X = 4.43 g of Oxygen
Mg+Cl2--> MgCl2
Magnesium plus chlorine equals magnesium chloride
In order to determine the mole ratio, you need to begin with a balanced chemical equation.
A0 = 1
a1 = 2
a2 = 1
a3 = 2
This can be solved by guessing and checking and making sure that all atoms on the left side are accounted for on the right side. Both sides must have the same amount of atoms or the balancing is not correct.
