For the reaction, A+2B→C,5 mol of A and 8 moles of B will produce:
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a
5 mole of C
b
4 mole of C
c
8 mole of C
d
13 mole of C
answer is B.
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Detailed Solution
A + 2B → C 5mol 8 mol 0Moles present 5mol 4mol 0 [∵2B=8, so B=4]Moles left 5−4 4−4 4 =1 =0[Bis a limiting reactant because its moles (=4) are less than those of A(=5)].