$5 A_{2} + 2 B_{4} \to 2 A B_{2} + 4 A_{2} B$
The minimum mass of mixture of $A_{2}$ and $B_{4}$ in grams required to produce at least $1 kg$ of each product is:
(Given atomic mass of $^{'} A^{'} = 10$ ; Atomic mass of $^{'} B^{'} = 120$ )

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Solution

$5 A_{2} + 2 B_{4} \to 2 A B_{2} + 4 A_{2} B$
Molar mass of $A_{2} = 20 g/mol$
molar mass of $B_{4} = 480 g/mol$
molar mass of $A B_{2} = 250 g/mol$
molar mass of $A_{2} B = 140 g/mol$
mass of $A B_{2} = 1000 g$
mass of $A_{2} B = 1000 g$
moles of $A B_{2} = \frac{1000}{250} = 4 mol$
moles of $A_{2} B = \frac{1000}{140} = 7.14 mol$

$\frac{moles of A B_{2}}{stoichiometric coefficient} = \frac{4}{2} = 2$
$\frac{moles of A_{2} B}{stoichiometric coefficient} = \frac{7.14}{4} = 1.785$

In the case of the product, limiting product is the one that has the maximum ratio.

Here limiting product is $A B_{2}$
Moles required of $A_{2}$ $= \frac{5 \times 4}{2} = 10$
So, mass of $A_{2}$ needed $= 10 \times 20 = 200 gm$
moles of $B_{4}$ needed $= \frac{2 \times 4}{2} = 4$
mass of $B_{4}$ required = $4 \times 480 = 1920$ Total mass of mixture $= 2120 g$

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