Question

$5A_2+2B_4\to 2A{B}_2+4A_{2}B$

The minimum mass of mixture of $A_2$ and $B_4$ required to produce at least $1$ kg of each product is :

(Given atomic mass of $A=10,B=120$)

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

Answer 1

The correct option is A $2120$ g

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 $10$ moles of $A_2$ and $4$ moles of $B_4$ respectively.
$1$ kg of $A_{2}B$ corresponds to $7.14$ moles.
It requires $8.9$ moles $A_2$ and $3.57$ moles $B_4$ respectively.
Thus, $10$ moles $A_2$ ($200$ g) and $4$ moles $B_4$ ($1920$ g) will produce atleast $1$ kg of each product.
The minimum mass of mixture required is $200$ g$+1920$ g $=2120$ g.