Initial concentration of magnesium nitrate M1 = 2.13 M
Initial volume of magnesium nitrate, MgNO3 V1 = 1.24 L
Final concentration of MgNO3, M2 = 1.60 M
Let the final volume of MgNO3 upon dilution be V2
Formula to use:
M1*V1 = M2*V2
V2 = M1*V1/M2
= 2.13 M * 1.24 L/1.60 M = 1.65 L
Thus, the final volume of magnesium nitrate solution upon dilution is 1.65 L
Answer:
10.1 g of Al are formed
Explanation:
The reaction is:
2AlCl3 --> 2Al + 3Cl2
So 2 moles of aluminun chloride decompose into 2 moles of Al and 3 moles of chlorine.
Ratio is 2:2.
Let's convert the mass of salt into moles (mass / molar mass)
50 g / 133.34 g/mol = 0.374 moles.
As the ratio is 2: 2, if I have 0.374 moles of salt, I would produce the same amount of Al, just 0.374.
Let's convert the moles to mass
(Mol . molar mass)
0.374 mol . 26.98 g / mol = 10.1 g of Al are formed
Ξ
