Home
Class 12
CHEMISTRY
In the reaction As(2)S(5)+xHNO(3)rarr5H(...

In the reaction `As_(2)S_(5)+xHNO_(3)rarr5H_(2)SO_(4)+yNO_(2)+2H_(3)AsO_(4)`
`+12H_(2)O`. The values of x and y are

A

40, 40

B

10, 10

C

30, 30

D

20, 20

Text Solution

AI Generated Solution

The correct Answer is:
To find the values of \( x \) and \( y \) in the reaction \[ \text{As}_2\text{S}_5 + x \text{HNO}_3 \rightarrow 5 \text{H}_2\text{SO}_4 + y \text{NO}_2 + 2 \text{H}_3\text{AsO}_4 + 12 \text{H}_2\text{O} \] we will balance the equation step by step. ### Step 1: Identify the elements involved In the reaction, we have the following elements: - Arsenic (As) - Sulfur (S) - Hydrogen (H) - Nitrogen (N) - Oxygen (O) ### Step 2: Count the number of atoms on the left side From the reactants: - Arsenic: 2 (from \(\text{As}_2\text{S}_5\)) - Sulfur: 5 (from \(\text{As}_2\text{S}_5\)) - Hydrogen: \( x \) (from \(\text{HNO}_3\)) - Nitrogen: \( x \) (from \(\text{HNO}_3\)) - Oxygen: \( 3x \) (from \(\text{HNO}_3\)) ### Step 3: Count the number of atoms on the right side From the products: - Arsenic: 2 (from \(2 \text{H}_3\text{AsO}_4\)) - Sulfur: 5 (from \(5 \text{H}_2\text{SO}_4\)) - Hydrogen: \( 10 + 6 + 24 = 40 \) (10 from \(5 \text{H}_2\text{SO}_4\), 6 from \(2 \text{H}_3\text{AsO}_4\), and 24 from \(12 \text{H}_2\text{O}\)) - Nitrogen: \( y \) (from \(y \text{NO}_2\)) - Oxygen: \( 20 + 2y + 12 = 32 + 2y \) (20 from \(5 \text{H}_2\text{SO}_4\), \(2y\) from \(y \text{NO}_2\), and 12 from \(12 \text{H}_2\text{O}\)) ### Step 4: Set up equations based on the balanced atoms From the hydrogen balance: \[ x = 40 \] From the nitrogen balance: \[ y = x = 40 \] From the oxygen balance: \[ 3x = 32 + 2y \] ### Step 5: Substitute \( x \) and \( y \) into the oxygen balance equation Substituting \( x = 40 \) and \( y = 40 \): \[ 3(40) = 32 + 2(40) \] \[ 120 = 32 + 80 \] \[ 120 = 112 \quad \text{(This is incorrect, indicating a need to check the calculations)} \] ### Step 6: Re-evaluate the oxygen balance Revisiting the oxygen balance: \[ 3x = 32 + 2y \] Substituting \( x = 40 \): \[ 3(40) = 32 + 2y \] \[ 120 = 32 + 2y \] \[ 2y = 120 - 32 \] \[ 2y = 88 \] \[ y = 44 \] ### Final Answer Thus, the values of \( x \) and \( y \) are: \[ x = 40, \quad y = 44 \]

To find the values of \( x \) and \( y \) in the reaction \[ \text{As}_2\text{S}_5 + x \text{HNO}_3 \rightarrow 5 \text{H}_2\text{SO}_4 + y \text{NO}_2 + 2 \text{H}_3\text{AsO}_4 + 12 \text{H}_2\text{O} \] we will balance the equation step by step. ...
Promotional Banner

Similar Questions

Explore conceptually related problems

In the reaction. H_(2)S+H_(2)O_(2) rarr S+2H_(2)O describes

In the reaction MnO_(4)^(-)+SO_(3)^(-2)+H^(+)rarrSO_(4)^(-2)+Mn^(2+)+H_(2)O

In the reaction MnO_(4)^(-)+SO_(3)^(-2)+H^(+)rarrSO_(4)^(-2)+Mn^(2+)+H_(2)O

In the following reaction As_(2)S_(3)+NO_(3)^(ɵ)+H_(2)OtoAsO_(4)^(3-)+SO_(4)^(2-)+NO+H^(o+) The number of electrons involved in the oxidation reaction is

Is the reaction BaO_(2)+H_(2)SO_(4)rarrBaSO_(4)+H_(2)O_(2) a redox reaction?

In the reaction CrO_(5) + H_(2)SO_(4)rarr Cr_(2)(sO_(4))_(3)+H_(2)O+O_(2) , one mole of CrO_(5) will liberate how many molesof O_(2) :

H_(2)underline(S)O_(5)+H_(2)O to H_(2)SO_(4)+H_(2)O_(2)

H_(2)underline(S)O_(5)+H_(2)O to H_(2)SO_(4)+H_(2)O_(2)

H_(2)underline(S_(2))O_(8)+H_(2)O to H_(2)SO_(4)+H_(2)O_(2)

H_(2)underline(S_(2))O_(8)+H_(2)O to H_(2)SO_(4)+H_(2)O_(2)