Question

Sodium nitrate decomposes according to the equation: 2NaNO3(s) = 2NaNO2(s) +O2(g)Calculate the volume (in cm3) of oxygen produced (measured at STP and at RTP) when 0.820g of sodium nitrate decomposes

Answer 1

It's simple task. Good luck!

Answer 2

First, we need to calculate the number of moles of sodium nitrate (NaNO3) present in 0.820g of the compound.

The molar mass of NaNO3 is: Na: 22.99 g/mol N: 14.01 g/mol O: 16.00 g/mol

So, the molar mass of NaNO3 is: 22.99 + 14.01 + (3 * 16.00) = 85.00 g/mol

Now, we can calculate the number of moles of NaNO3: 0.820g / 85.00 g/mol = 0.00965 moles

According to the balanced chemical equation, 2 moles of NaNO3 produce 1 mole of O2. Therefore, 0.00965 moles of NaNO3 will produce: 0.00965 moles * (1 mole O2 / 2 moles NaNO3) = 0.004825 moles of O2

Now, we can use the ideal gas law to calculate the volume of O2 produced at STP (Standard Temperature and Pressure) and RTP (Room Temperature and Pressure).

At STP (0°C and 1 atm), 1 mole of any gas occupies 22.4 L. Therefore, the volume of O2 produced at STP is: 0.004825 moles * 22.4 L/mol = 0.108 L = 108 cm3

At RTP (20°C and 1 atm), the molar volume of a gas is approximately 24 L/mol. Therefore, the volume of O2 produced at RTP is: 0.004825 moles * 24 L/mol = 0.116 L = 116 cm3

So, the volume of oxygen produced (measured at STP) when 0.820g of sodium nitrate decomposes is 108 cm3, and the volume of oxygen produced (measured at RTP) is 116 cm3.