Respuesta :
We can calculate for the number of moles of NO2 using its molar mass 46.0055 g/mol:
11.5 g NO2 (1 mol NO2 / 46.0055 g NO2) = 0.250 mol NO2
We use the mole ratio of O2 and NO2 from their coefficients in the balanced chemical equation
2Pb(NO3)2(s) → 2PbO(s) + 4NO2(g) + O2 (g)
which is one mole O2 is to react with four moles of NO2, to compute for the number of moles of oxygen:
0.250 mol NO2 (1 mol O2 / 4 mol NO2) = 0.0625 mol O2
We can now calculate for the mass of oxygen O2 using its molar mass:
0.0625 mol O2 (31.998 g O2 / 1 mol O2) = 2.0 grams O2
Therefore, 2.0 grams of oxygen is produced.
11.5 g NO2 (1 mol NO2 / 46.0055 g NO2) = 0.250 mol NO2
We use the mole ratio of O2 and NO2 from their coefficients in the balanced chemical equation
2Pb(NO3)2(s) → 2PbO(s) + 4NO2(g) + O2 (g)
which is one mole O2 is to react with four moles of NO2, to compute for the number of moles of oxygen:
0.250 mol NO2 (1 mol O2 / 4 mol NO2) = 0.0625 mol O2
We can now calculate for the mass of oxygen O2 using its molar mass:
0.0625 mol O2 (31.998 g O2 / 1 mol O2) = 2.0 grams O2
Therefore, 2.0 grams of oxygen is produced.
The mass of oxygen formed during the decomposition of lead nitrate is 2 grams.
The complete balanced equation will be:
[tex]\rm 2\;Pb(NO_3)_2\;\rightarrow\;2\;PbO\;+\;4\;NO_2\;+\;O_2[/tex]
From the reaction the decomposition of 2 moles of lead nitrate will give 4 moles of Nitric oxide and 1 mole of oxygen
The ratio of [tex]\rm NO_2[/tex] : Oxygen has been 2 : 1.
The moles of [tex]\rm NO_2[/tex] formed:
Moles = [tex]\rm \dfrac{weight}{molecular\;weight}[/tex]
Molecular weight of [tex]\rm NO_2[/tex] = 46 g/mol
Moles of [tex]\rm NO_2[/tex] = [tex]\rm \dfrac{11.5}{46}[/tex]
Moles of [tex]\rm NO_2[/tex] = 0.25 moles.
The moles of [tex]\rm NO_2[/tex] = [tex]\rm \dfrac{1}{4}[/tex] moles of Oxygen
Moles of oxygen = [tex]\rm \dfrac{0.25}{4}[/tex]
Moles of oxygen = 0.06 mol.
The molecular weight of Oxygen = 32g/mol
Mass of 0.06 mol of Oxygen = 32 [tex]\times[/tex] 0.06
Mass of Oxygen = 2 grams.
The mass of oxygen formed during the decomposition of lead nitrate is 2 grams.
For more information about mass of substance formed, refer to the link:
https://brainly.com/question/21183464
