the mole Watch
What volume of ammonia gas will be produced in the following reaction if 50 cm3 of 0.5 mol/dm3 sodium hydroxide is boiled with 50 cm3 of 0.4 mol/dm3 ammonium chloride solution?
(Note one of these is in excess)
NaOH (aq)+NH4CI (aq)–> Nacl (aq) +H2O (l) NH3(g)
So first convert the volume of NH4Cl solution from centimetres cubed into decimetres cubed, given there are 1000 cm3 per dm3.
Then multiply the number of dm3 of NH4Cl solution by the concentration of the NH4Cl solution, which is in mol/dm3. (The dm3 and /dm3 cancel and give an amount of NH4Cl, in moles.)
The formula tells us NH4Cl is converted into NH3 in a 1:1 ratio, so we have THAT many moles of NH3. (I.e. the same number of moles of NH3 as we had NH4Cl.)
At standard conditions, a mole of gas takes up 22.4L of volume. So multiply the amount of NH3 we had by 22.4 to give the volume of NH3 in litres. If this is an awkward number to write down you may want to convert it to mL.