mod 6 MC (1 Viewer)

[NexusRich]

please explain how to do this

 

[username_2]

please explain how to do this

So the total mass of the sample is 4.52g which will obviously be in the denominator because we are finding the %weight/weight ratio between sulphate and the original sample.

Now to determine the numerator, we need to know how much of the barium sulphate mass is actually sulphate.

Consider the equation: Ba + SO4 -> BaSO4

So here the molar ratio between the SO4 and the BaSO4 is 1:1, hence if BaSO4 if n moles, then SO4 will also have n moles

To find n we use n = m(BaSO4)/MM(BaSO4) = (3.62g)/(233.39 g/mol) = 0.01551... mol = n(SO4)

Using the same equation, we can determine m(SO4) = n(SO4) * MM(SO4) = 0.01551... mol * 96.07 g/mol = 1.49... g

Therefore the %w/w would be m(SO4)/m(BaSO4) = 1.49.../4.52 * 100 = 32.97... % ~ 33.0% = B

Hence, the answer. Hope it helps