You visited us 1 times! Enjoying our articles? Unlock Full Access!

Calcination

If the supply...

Question

If the supply of oxygen is limited, $H_{2} S$ reacts with $O_{2}$ to form:

H_{2} O + S O_{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

H_{2} O + S

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

H_{2} {S O}_{4} + S

No worries! We‘ve got your back. Try BYJU‘S free classes today!

H_{2} O + S O_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $H_{2} O + S$
In the limited supply of $O_{2}, H_{2} S$ burns with a blue flame. $S$ and $H_{2} O$ are the main products.

$2 H_{2} S + O_{2} ⟶ 2 H_{2} O + 2 S$

Suggest Corrections

Similar questions

H_{2} S + \frac{3}{2} O_{2} \to H_{2} O + S O_{2},

Δ H = - 136.7 K c a l

H_{2} + \frac{1}{2} O_{2} \to H_{2} O, Δ H = - 68.4 K c a l

S + O_{2} \to S O_{2}, Δ H = - 71 K c a l

Δ H_{f} o f H_{2} S

S + O_{2} \to {S O}_{2}; △ H = - 298.2 k J

{S 0}_{2} + 1 / 20_{2} \to {S O}_{3}; △ H = - 98.7 k J

S O_{3} + H_{2} O \to H_{2} {S O}_{4}; △ H = - 130.2 k J

H_{2} + 1 / 20_{2} \to H_{2} O; △ H = - 287 k J

H_{2} {S O}_{4}

C u + H_{2} {S O}_{4} - \to C u S O_{4} + {S O}_{2} ↑ + H_{2} O

F e + + H_{2} O - \to {F e}_{3} O_{4} + H_{2} ↑

{C H}_{4} + O_{2} \to C O_{2} + H_{2} O

S + O_{2} \to {S O}_{2}; Δ H = - 298.2 k J

{S O}_{2} + \frac{1}{2} O_{2} \to S O_{3}; Δ H = - 98.7 k J

{S O}_{3} + H_{2} O \to H_{2} {S O}_{4}; Δ H = - 130.2 k J

H_{2} + \frac{1}{2} O_{2} \to H_{2} O; Δ H = - 227.3 k J

H_{2} {S O}_{4}

H_{2} S

H_{2} O

E_{H^{+} / O_{2} / P t}^{o} > E_{H^{+} / H_{2} S / S}^{o}

Join BYJU'S Learning Program