What volume of nitrogen dioxide is produced from the complete reaction of 16.87g of lead at STP?

2Pb(NO3)——> 2PbO(s)+4NO2(g)+O2(g)

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superman1987 superman1987 · Answer 1 · 2019-01-03T23:03:45+01:00

Answer: 16.78 g of lead(II) nitrate will produce 2.238 L of NO2

Explanation:

From the balanced equation, it is clear that every mole of lead nitrate will produce 2 moles of NO2.
n(no. of moles) = mass/molar mass

n(lead nitrate) = 16.87/331.2 = 0.05 mole

0.05 mole of lead nitrate will produce 0.1 mole of nitrogen dioxide (NO2)

at STP: P = 1.0 atm and T = 273.0 K

R = 0.082 L.atm/mol.K

V = nRT/P

V = (0.10 mol)(0.082 L.atm/mol.K)(273.0 K)/(1.0 atm) = 2.238 L

CeaciliaTarifil CeaciliaTarifil · Answer 2 · 2021-03-11T20:38:08+01:00

Answer: 2.28 mol No2

2Pb(No3)2->2Pbo+4No2+O2

Explanation:

1. List known factors

-16.87g Pb(No3)2

-mol mas Pb(No3)2=331.2

-2Pb(No3)2=4No2<---This is your ratio of (Pbno3) to (No2)

-1 mol gas at STP is 22.4, this'll help finish your equation out. This remains the same for all equations.

2.Solve

[tex]16.87g Pb(No3)2*\frac{1 mol Pb(no3)2}{331.2} *\frac{4 NO2}{2Pb(No3)2}*22.4L[/tex]

Hope this helps you in the future! I sure had a time figuring it out myself, heehee!

-Michiko