Answer:
The formula of the compound is:
N2H2
Explanation:
Data obtained from the question:
Nitrogen (N) = 93.28%
Hydrogen (H) = 6.72%
Next, we shall determine the empirical formula for the unknown compound. This is illustrated below:
N = 93.28%
H = 6.72%
Divide by their molar mass
N = 93.28 /14 = 6.663
H = 6.72 /1 = 6.7
Divide by the smallest
N = 6.663 / 6.663 = 1
H = 6.72 /6.663 = 1
Therefore, the empirical formula is NH.
Now, we can obtain the formula of the compound as follow:
The formula of a compound is simply a multiple of the empirical formula.
[NH]n = 30.04
[14 + 1]n = 30.04
15n = 30.04
Divide both side by 15
n = 30.04/15
n = 2
Therefore, the formula of the compound is:
[NH]n => [NH]2 => N2H2
The first one will be 2 1 -> 2 1
Second is 2 3 -> 6 1
If concentration of HCl is 1 mol/dm³ :
m(<span>erlenmeyer flask) = 88,00 g.
m(Zn) = 25,0 g.
V(HCl) = 15 ml = 15 cm</span>³ = 0,015 dm³.
Chemical reaction: Zn + 2HCl → ZnCl₂ + H₂.
n(HCl) = c(HCl) · V(HCl).
n(HCl) = 1 mol/dm³ · 0,015 dm³ = 0,015 dm³.
n(Zn) = 25 g ÷ 65,4 g/mol = 0,38 mol.
n(H₂) = 0,015 mol ÷ 2 = 0,0075 mol.
m(H₂) = 0,0075 mol · 2g/mol = 0,015 g.
