Question

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Answer 1

Answer:

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Answer 2

Solution :-  

Given chemical equation to balance :-  

$\small{\sf{As_2S_3 + HNO_3 + H_2O \rightarrow H_3AsO_4 + H_2SO_4 + NO}}$  $\small{\sf{As_{2}^{+3}S_{3}^{-2} + HN^{+5}O_3 + H_2O \rightarrow H_3As^{+5}O_4 + H_2S^{+6}O_4 + N^{+2}O}}$

 Now Change in Oxidation Number  

$\bigstar \sf{ As^{+3} \rightarrow As^{+5} \: , \Delta change = 2 }$

Oxidation  

$\bigstar \sf{ S^{-2} \rightarrow S^{+6} \: , \Delta change = 8 }$  Oxidation

 $\bigstar \sf{ N^{+5} \rightarrow N^{+2} \: , \Delta change = 3 }$  Reduction  

So as we know that we cross multiply the reduction and the Oxidation part , but as from same compound are oxidising , So we will multiply the oxdised part by the change of reduction part and balance further.  

So our partially balanced equation :-  

$= \small{\sf{(3)As_2S_3 + HNO_3 + H_2O \rightarrow (2)(3)H_3AsO_4 + (3)(3)H_2SO_4 + NO}}$

 $= \small{\sf{3As_2S_3 + HNO_3 + H_2O \rightarrow 6H_3AsO_4 + 9H_2SO_4 + NO}}$  

Now we can see that the no. of Hydrogen is fixed but still Nitrogen isn't balanced .  

We will assign variables "x" and "y" at respective places  in order to balance.  

$= \small{\sf{3As_2S_3 + xHNO_3 + yH_2O \rightarrow 6H_3AsO_4 + 9H_2SO_4 + xNO}}$  

So while writing for Hydrogen  

→ x + 2y = 6(3) + 9(2)  

→ x + 2y = 18 + 18  

→ x + 2y = 36 .....(i)

Writing for Oxygen  

→ 3x + y = 6(4) + 9(4) + x  

→ 2x + y  = 24 + 36  

→ 2x + y = 60 ......(ii)  

Now by 2×(ii) - (i)

4x + 2y = 120

 (-)x ± 2y = (-)36

 _________________

3x = 84  

→ x = 28  

Now by putting the value of "x" in (i)

→ x + 2y = 36  

→ 28 + 2y = 36  

→ 2y = 8  

→ y = 4  

So now we will replace the values of "x" and "y" in the equation.  

So our balanced equation :-  

$= \small{\sf{3As_2S_3 + 28HNO_3 + 4H_2O \rightarrow 6H_3AsO_4 + 9H_2SO_4 + 28NO}}$  

Now checking the equation :-  

$\begin{array}{| c | c | c |}\cline{1-3} Atom & In\: LHS & In \: RHS \\\cline{1-3} As & (2)(3) = 6 & (6)(1) = 6 \\\cline{1-3} S & (3)(3) = 9 & (9)(1) = 9 \\\cline{1-3} H & 28 + (4)(2) = 36 & (6)(3) + (9)(3) = 36 \\ \cline{1-3} N & 28 & 28 \\\cline{1-3} O & 28(3) + 4 = 88 & 6(4) + 9(4) + 28 = 88 \\ \cline{1-3}\end{array}$  

Hence Balanced