Question

What volume of nitrogen dioxide (in L) is produced from the complete reaction of 16.87g of lead(II)nitrate at STP
2Pb(NO3)2 (g) -->2PbO(s) + 4NO2(g) + O2

Answer 1

Use stoichiometry with the conversion that 1 mol of gas at STP = 22.4 L of that gas

Answer 2

We know we completely react 16.87g of our lead nitrate reactant. So, in order to utilise our balanced chemical equation to find out how much product we form, we first need to convert this into moles:

\[No.~of~moles = \frac{ No.~of~grams }{ Molecular~mass~(g/mol) }\]

The molecular mass of lead nitrate is 331.2 g/mol, which you can get from adding up the atomic weights of all the atoms that form part of the molecule (you'll find these on the periodic table!)

\[No.~of~moles = \frac{ 16.87~g }{ 331.2~g/mol } = x\]

From our balanced chemical equation, we see that there is a '2' in front of the lead nitrate and a '4' in front of the nitrogen dioxide (the product we're interested in). This tells us that for every 2 moles of lead nitrate which we react fully, we will form 4 moles of NO2 gas.

So, the molar ratio of lead nitrate to nitrogen dioxide is 2:4, or in simpler terms 1:2. This tells us that we will form twice the number of moles of NO2 than lead nitrate which we reacted. So, if we have x moles of lead nitrate, we will form 2x moles of NO2 as our product.

We now need to convert this value into litres. As @JoannaBlackwelder said, at standard temperature and pressure 1 mole of a gas will occupy 22.4 litres (at room temperature it's 24 litres).

\[1~mole = 22.4~litres\]
\[2x~moles = (2x)(22.4)~litres\]
which should give you the answer! Hopefully this provides you with a pathway to tackle the problem @mmaarissaa :)