Question

How many grams of $90$% pure $N{a}_{2}S{O}_3$ can be produced from $250g$ of $95$% pure $NaCl$?

• A. $640.6g$
• B. $288.2g$
• C. $259.4g$
• D. $320.3g$

Answer 1

The correct option is A $640.6g$

Given $90$% pure $N{a}_{2}S{O}_3$ is produced from $250g$ $95$% pure $NaCl$

The reaction involved is

$2NaCl+H_{2}S{O}_3⟶2HCl+N{a}_{2}S{O}_3$

We know actual mass of $NaCl=95$% of $250g$

$=\frac{95}{100}\times 250$

$=237.5g$

From above reaction , it is clear that

$2$ moles of $NaCl⟶1$ mole $N{a}_{2}S{O}_3$

i.e. $117g$ $NaCl⟶126g$ $N{a}_{2}S{O}_3$

$\therefore 237.5g$ $NaCl⟶Xg$ $N{a}_{2}S{O}_3$

$\therefore x=\frac{237.5\times 126}{117}g$

$=255.76g$ $N{a}_{2}S{O}_3$ is produced.

But this $N{a}_{2}S{O}_3$ is $100$% pure,

So $90$% pure $N{a}_{2}S{O}_3$ produced

$=\frac{90}{100}\times 255.76g$

$=230.184g$

$\therefore 230.184g$ $90$% $N{a}_{2}S{O}_3$ is produced