Question

The equation below shows the decomposition of lead nitrate. How many grams of oxygen are produced when 11.5 g NO is formed?

2 Pb(NO3)2 (s) = 2 PbO (s) + 4 NO2 (g) + O2 (g)

Answer 1

To find the grams of oxygen produced in the decomposition of lead nitrate when 11.5 g of NO is formed, use stoichiometry. The calculation would be 11.5 g NO * (1 mol NO / 30.01 g NO) * (1 mol O2 / 2 mol NO) * (32 g O2 / 1 mol O2) = 19.2 g O2.

Step-by-step explanation:

To find the number of grams of oxygen produced when 11.5 g of NO is formed in the decomposition of lead nitrate, we need to use stoichiometry. According to the balanced chemical equation, 1 mole of O2 is produced for every 2 moles of Pb(NO3)2 decomposed. The molar mass of O2 is 32 g/mol.

First, calculate the moles of NO produced using the given mass of NO and its molar mass. Then, use the mole ratio between NO and O2 to calculate the moles of O2. Finally, multiply the moles of O2 by its molar mass to find the mass in grams.

In this case, the calculation would be:

11.5 g NO * (1 mol NO / 30.01 g NO) * (1 mol O2 / 2 mol NO) * (32 g O2 / 1 mol O2) = 19.2 g O2