# the moleWatch

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#1
Pls pls help with full detailed working

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)
0
3 weeks ago
#2
Are we told anything about the conditions? are they standard 298K 100kPa?
0
#3
no.. that is the full qn.
0
3 weeks ago
#4
the rest is en route, but does this look okay?
0
3 weeks ago
#5
Nevermind that's all wrong
0
3 weeks ago
#6
(Original post by bj48394)
the rest is en route, but does this look okay?
Be careful not to just give her the answers. We dont do that here.
0
3 weeks ago
#7
Yeah but the working explains hopefully
0
3 weeks ago
#8
There's more NaOH there than NH4Cl, and the equation is 1:1 with respect to those reactants, so the NaOH is in excess and the NH4Cl is the limiting ingredient.

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.
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