Chemistry, 24.07.2019 18:00 potato3999

The equation below shows the decomposition of lead nitrate. how many grams of oxygen are produced when 11.5 g no2 is formed? 2pb(no3) 2(s) --> 2pbo(s) + 4no2 (g) + o2(g

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Answer from: stina2021

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.

Answer from: yselahernandez02

The mass of oxygen formed during the decomposition of lead nitrate is 2 grams.

The complete balanced equation will be:

$\rm 2\;Pb(NO_3)_2\;\rightarrow\;2\;PbO\;+\;4\;NO_2\;+\;O_2$

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 $\rm NO_2$ : Oxygen has been 2 : 1.

The moles of $\rm NO_2$ formed:

Moles = $\rm \dfrac{weight}{molecular\;weight}$

Molecular weight of $\rm NO_2$ = 46 g/mol

Moles of $\rm NO_2$ = $\rm \dfrac{11.5}{46}$

Moles of $\rm NO_2$ = 0.25 moles.

The moles of $\rm NO_2$ = $\rm \dfrac{1}{4}$ moles of Oxygen

Moles of oxygen = $\rm \dfrac{0.25}{4}$

Moles of oxygen = 0.06 mol.

The molecular weight of Oxygen = 32g/mol

Mass of 0.06 mol of Oxygen = 32 $\times$ 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:

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Sorry i dont understand the question can you rephase it?

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