2 moles of H 2 S and 11.2 L of SO 2 at N.T.P. reacts to form x moles of sulphur. The value of x is:

Question

2 moles of H 2 S and 11.2 L of SO 2 at N.T.P. reacts to form x moles of sulphur. The value of x is: (A) 1.5 (B) 3.5 (C) 7.8 (D) 12.7

Tardigrade Admin · Accepted Answer

22.4 L of SO 2 at N ⋅ T.P =1 moles 11.2 L of SO 2 at N . T.P =1 / 2 moles SO 2+2 H 2 S arrow 3 S +2 H 2 O Here SO 2 is a limiting reactant ∴ moles of sulphur produced =3 × moles of SO 2 =1.5 moles

Q. 2 moles of $H_{2} S$ and $11.2 L$ of $S O_{2}$ at $N . T . P .$ reacts to form x moles of sulphur. The value of $x$ is:

1.5

3.5

7.8

12.7

$22.4 L$ of $S O_{2}$ at $N \cdot T . P = 1$ moles

$11.2 L$ of $S O_{2}$ at $N . T . P = 1/2$ moles

$S O_{2} + 2 H_{2} S \to 3 S + 2 H_{2} O$

Here $S O_{2}$ is a limiting reactant

$∴$ moles of sulphur produced $= 3 \times$ moles of $S O_{2}$

$= 1.5$ moles

Q. 2 moles of H2​S and 11.2L of SO2​ at N.T.P. reacts to form x moles of sulphur. The value of x is:

1.5

3.5

7.8

12.7

22.4L of SO2​ at N⋅T.P=1 moles 11.2L of SO2​ at N.T.P=1/2 moles SO2​+2H2​S→3S+2H2​O Here SO2​ is a limiting reactant ∴ moles of sulphur produced =3× moles of SO2​ =1.5 moles

Q. 2 moles of $H_{2} S$ and $11.2 L$ of $S O_{2}$ at $N . T . P .$ reacts to form x moles of sulphur. The value of $x$ is:

$22.4 L$ of $S O_{2}$ at $N \cdot T . P = 1$ moles

$11.2 L$ of $S O_{2}$ at $N . T . P = 1/2$ moles

$S O_{2} + 2 H_{2} S \to 3 S + 2 H_{2} O$

Here $S O_{2}$ is a limiting reactant

$∴$ moles of sulphur produced $= 3 \times$ moles of $S O_{2}$

$= 1.5$ moles