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Consider the reaction
$5 A_{2} + 2 B_{4} \to 2 A B_{2} + 4 A_{2} B$
Atomic weight of ‘A’ = 10 &Atomic weight of ‘B’ = 120 then the correct statement(s) is (are)

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The minimum mass of mixture of $A_{2}$ and $B_{4}$ required to produce at least 1Kg of each product is 2120 g

The minimum mass of mixture of $A_{2}$ and $B_{4}$ required to produce atleast 250g of each product is 730 g

If charge of ‘B’ on product side is -2, then the n-factor of $A_{2}$ for the given reaction is 16/5

If charge of ‘B’ on product side is -2, then the n-factor of $B_{4}$ for the above reaction is ‘2’

answer is A, C.

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Detailed Solution

$10 \times 20 = 200 \begin{array}{l} 4 \times 480 = 1920 \\ \Rightarrow 2120 gm \end{array}$

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