My suggested answer:
Q21) D
Q22) C
Q23) D
Q24) C
Q25) A
Q26) B
Out of the 20 chemistry question, the question I like most will be
this. On the surface it looked simple, but it's testing Limiting
reactant, and calculation of excess reactant.
First step is to determine the molecular formula. After you've done
with your calculation, you'll get FeS.
Then next is to find out which is the the limiting reactant. After
the calculation, you'll know that Fe is the limiting reactant, and
of course S is in excess.
No. of mol of FeS/ No. of mol of Fe = 1/1 = 1
No. of mol of FeS = 1 x 0.125 = 0.125 mol
Mass of FeS = 0.125 x (56 + 32) = 11g
Final step is to fine out the mass of excess reactant(S).
As calculated earlier(not shown here), take the no. of mole of S
minus the no. of moles of Fe. Which is 0.219 - 0.125 = 0.094 moles
of S in excess
Then 0.094 x 32(Ar of S) = 3.
Answer: B
Q27) D
Q28 ) B
Q29) D
Q30) C
Q31) A
Q32) D
Q33) A
Q34) C
Q35) C
Q36) C
Q37) A
Q38 ) B
Q39) A
Q40) B
Edited for Q34, thanks kester for pointing out my mistake.
