When Dinitrogen Pentoxide is heated, it decomposes to nitrogen dioxide and oxygen. Find The Volume Of NO2 and O2 gases produced at S.T.P when A Sample of 54g of N2O5 is heated.
Answers
Answer:
the balanced chemical equation for this answer is:
2 N2O5_________ 4NO2 + O2
first, we find no. of moles of N205
moles =mass divides by molar mass
molar mass of N205 is 276
54 divided by 108 =0.5 moles.
if 2 moles of N205 produce 4 moles of NO2
than 0.5 moles of N205 produce 1 moles of NO2
volume of N02 at S.T.P is 22.4x1
22.4 dm3 volume of NO2
if uh want to calculate at R.TP Simply you multiply 24x1 that is
22.4
you can find the volume of 02 by the same method.
2 moles of N2O5 produce 1 mole of 02
0.5 moles of N2O5 produce 0.25 moles
volume at S.T.P equals 22.4x0.25
5.6volume.
Explanation:
Answer:
Mass of O₂ = 0.5 × 32= 16g
Therefore the masses of nitrogen dioxide and oxygen are 92g and 16 g respectively.
Explanation:
When dinitrogen pentoxide, N2O5, a white solid, is heated, it decomposes to nitrogen dioxide and oxygen. 2N2O5(s) → 4NO2(g) + O2(g) Ifa pattern of N2O5 produces 1.381 g O2,what number of grams of NO2 are formed
Given
When dinitrogen pentoxide (N2O5) decomposes to nitrogen dioxide and oxygen.
to find
the quantity of NO₂ and O₂ gases produced while a pattern 54g of N₂O₅ is heated.
2N₂O₅ ⇒4NO₂ + O₂
Here moles of N₂O₅ decomposes into four moles of NO₂ and one mole of O₂.
mass of N₂O₅ = 54g
Molecular mass of N₂O₅ = 28 + 80 = 108g
⇒no of moles of N₂O₅ = 54/108 = 0.five
so, four × 0.five = 2 moles of NO₂ and 0.five mole of O₂ are produced
mass of NO₂ = 2 × 46 = ninety two g
Mass of O₂ = 0.five × 32= 16g
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