Question

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

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

Answer 1

The correct option is A $2120g$

Balanced chemical equation :
$5A_2+2B_4\to 2A{B}_2+4A_{2}B$
$\text{Molar mass of}A{B}_2=10+(2\times 120)=250g/mol$
$\text{Molar mass of}{A}_{2}B=(2\times 10)+120=140g/mol$

$\text{Number of moles of}A{B}_2=\frac{1000}{250}=4mol$
According to the stoichiometry of reaction, $2mol$ of $A{B}_2$ are produced by $5mol$ of $A_2$ and $2mol$ of $B_4$.
Therefore, $4mol$ of $A{B}_2$ are produced by $10mol$ of $A_2$ and $4mol$ of $B_4$.

$\text{Moles of}{A}_{2}B$ = $\frac{1000}{140}=7.14mol$
According to the stoichiometry of reaction, $4mol$ of $A{B}_2$ are produced by $5mol$ of $A_2$ and $2mol$ of $B_4$.
Therefore, $7.14mol$ of $A{B}_2$ are produced by $8.925mol$ of $A_2$ and $3.57mol$ of $B_4$.

Hence to produce $1kg$ of each product, we require $10mol$ of$A_2$ and $4mol$ of $B_4$.
Hence minimum mass of $A_2$ required is $20\times 10=200g$ and minimum mass of $B_4$ required is $4\times 480=1920g$.
Therefore, minimum masss required to produce $1kg$ of each product is $1920+200=2120g$.