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:
Higher pressure, is the right answer.
Explanation:
The A will have a higher pressure. Since we have given the volume and temperature is same in both containers A and B. Below is the calculation for proof that shows which container has the higher pressure while keeping the volume and temperature the same.
Therefore, the container “A” will have higher pressure.
Answer is: <span>the empirical formula of the hydrocarbon is CH</span>₂.<span>
Chemical reaction: C</span>ₓHₐ + O₂ → xC + a/2H₂O.<span>
m(CO</span>₂) = 33.01 g.
n(CO₂) = m(CO₂) ÷ M(CO₂).
n(CO₂) = 33.01 g ÷ 44.01 g/mol.
n(CO₂) = n(C) = 0.75 mol.
m(H₂O) = 13.52 g.
n(H₂O) = 13.52 g ÷ 18 g/mol.
n(H₂O) = 0.75 mol.
n(H) = 2 · n(H₂O) = 1.5 mol.
n(C) : n(H) = 0.75 mol : 1.5 mol /0.75 mol.
n(C) : n(H) = 1 : 2.
H2O
This equation is a double displacement reaction, and it forms H2CO3, which is very unstable and separates into H2O and CO2.
