Hi I was wondering if anyone could help me out with this question, it keeps going over my head.
Q) 20cm^3 of 0.2 mol l^-1 sulphuric acid is added to 50cm^3 of 0.1 mol l^-1 barium chloride solution.
a- calculate which reactant is in excess
b- what mass of barium sulphate would be precipitated?
I keep on getting the sulphuric acid as the excess and 1.167 g of product. the answer at the back says barium chloride is in excess and 0.934 g is produced. can someone please help me out?
any answers would be greatly appreciated and also how much studying have yous been doing for your highers?
limiting reagant Watch
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- 17-04-2014 18:42
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- 17-04-2014 19:43
I always start by writing out the chemical equation: H2SO4 + BaCl2 ---> BaSO4 + 2HCl
You need to work out the amounts in moles. From the equation above we can see that one mole of sulfuric acid reacts with one mole of barium chloride - easy peasy.
moles (mol) = volume (dm3) x concentration (mol dm-3)
H2SO4 = 0.020 x 0.2 = 0.004 mol
BaCl2 = 0.050 x 0.1 = 0.005 mol
There is more barium chloride than sulfuric acid, so it is in excess. Make sense?
We then use the limiting reagent (the one not in excess) to calculate the mass of the barium sulfate. Does that help?