Question

The minimum mass of the mixture of $A_2$ and $B_4$ required to produce at least $1kg$ of each product is:
(Given At. mass of $A=10$; At. mass of $B=120$)
$5A_2+2B_4⟶2A{B}_2+4A_{2}B$.

• A. $2120gm$
• B. $1060gm$
• C. $560gm$
• D. $1660gm$

Answer 1

The correct option is A $2120gm$

Molar masses of $A{B}_{2}and{A}_{2}B$ are 250 g/mol and 140 g/mole respectively.
1 kg of $A{B}_2$ corresponds to 4 moles. It requires 10 mole $A_2$ and 4 mole $B_4$ respectively.
1 kg of $A_{2}B$ corresponds to 7.14 mole. It requires 8.9 mole $A_2$ and 3.57 mole $B_4$ respectively.
Thus 10 mole $A_2$ (200 g) and 4 mole $B_4$ (1920 g) will produce atleast 1 kg of each product.
The minimum mass of mixture required is $200g+1920g=2120g$.