91 grams of sodium azide required to decompose and produce 2.104 moles of nitrogen.
Explanation:
2NaN3======2Na+3N2
This is the balanced equation for the decomposition and production of sodium azide required to produce nitrogen.
From the equation:
2 moles of NaNO3 will undergo decomposition to produce 3 moles of nitrogen.
In the question moles of nitrogen produced is given as 2.104 moles
so,
From the stoichiometry,
3N2/2NaN3=2.104/x
= 3/2=2.104/x
3x= 2*2.104
= 1.4 moles
So, 1.4 moles of sodium azide will be required to decompose to produce 2.104 moles of nitrogen.
From the formula
no of moles=mass/atomic mass
mass=no of moles*atomic mass
1.4*65
= 91 grams of sodium azide required to decompose and produce 2.104 moles of nitrogen.
Answer:
i. Lead nitrate:
2Pb (NO3)2 Δ= 2PbO+4NO2+O2
ii. Potassium chlorate:
2KClO3 → 2KCl + 3O2↑
iii. clacium carbonate:
CaCO3 + 2HCl -> CaCl2 + CO2 + H2O
iv. cupric carbonate :
CuCO3 → CuO + CO2↑
Hope this helped
All the best!!
First let's find out the oxidation number of Fe in K₄[Fe(CN)₆] compound.
The oxidation number of cation, K is +1. Hence, the total charge of the anion, [Fe(CN)₆] is -4. CN has charge has -1. There are 6 CN in anion. Let's assume the oxidation number of Fe is 'a'.
Sum of the oxidation numbers of each element = Charge of the compound
a + 6 x (-1) = -4
a -6 = -4
a = +2
Hence, oxidation number of Fe in [Fe(CN)₆]⁴⁻ is +2.
Now Fe has the atomic number as 26. Hence, number of electrons in Fe at ground state is 26.
Electron configuration = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s² = [Ar] 3d⁶ 4s²
When making Fe²⁺, Fe releases 2 electrons. Hence, the number of electrons in Fe²⁺ is 26 - 2 = 24.
Hence, the electron configuration of Fe²⁺ = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶
= [Ar] 3d⁶
Hence, the number of 3d electrons of Fe in K₄[Fe(CN)₆] compound is 6.
