The reactant which is entirely consumed in reaction is known as limiting reagent. In the reaction `2A+4Brarr3C+4D`, when 5 moles of A react with 6 moles of B, then
(a) which is the limiting reagent?
(b) calculate the amount of C formed?

Options

A. `A,2.5` moles of C
B. `B,4.5` moles of C
C. `C,4` moles of C
D. `D,3.5` moles of C

Answer

B

Explanation

`2A +4B rarr 3C +4D`, According to the equation. 2 moles of 'A' requite 4 moles of 'B' for the reaction. Hence, for 5 moles of 'A' the moles of 'B' required
= 5 mol of `A xx (4 "mol of" B)/(2 "mol of"A)`
But have hence only 6 moles of 'B' is the limiting reagent. So amount of 'C" formed is determined by amount of 'B'. Since 4 moles of 'B' give 3 moles of 'C'. hence 6 moles of 'B' will give
6 mol of `B xx (3 "mol of" C)/(4 "mol of" B) = 4.5` mol of C