At STP, 2.8 litres of hydrogen sulphide were mixed with 1.6 litres of sulphur dioxide and the reaction occurred according to the equation 2H2S(g)+SO2(g) arrow 2H2O(l)+3S(s) Which of the following shows the volume of the gas remaining after the reaction?

Question

At STP, 2.8 litres of hydrogen sulphide were mixed with 1.6 litres of sulphur dioxide and the reaction occurred according to the equation 2H2S(g)+SO2(g) arrow 2H2O(l)+3S(s) Which of the following shows the volume of the gas remaining after the reaction? (A) 0.2 litres of SO2(g) (B) 0.4 litres of H2(g) (C) 1.2 litres of H2S(g) (D) 1.2 litres of SO2(g)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/c-kvvctlxy7bnd.png" /> So, textSol text2 textvol textH text2 textS ≡ text1 textvol textSO text2 or text2 text.8 textL textH text2 textS text= text1 text.4 textL textSO text2 Hence, after reaction, volume of remaining SO2 = 1.6 - 1.4 textL = 0.2 textL

Q. At STP, 2.8 litres of hydrogen sulphide were mixed with 1.6 litres of sulphur dioxide and the reaction occurred according to the equation

$2 H_{2} S (g) + S O_{2} (g) \to 2 H_{2} O (l) + 3 S (s)$

Which of the following shows the volume of the gas remaining after the reaction?

0.2 litres of $S O_{2} (g)$

0.4 litres of $H_{2} (g)$

1.2 litres of $H_{2} S (g)$

1.2 litres of $S O_{2} (g)$

So, $Sol 2 vol H_{2} S \equiv 1 vol SO_{2}$

or $2 .8 L H_{2} S = 1 .4 L SO_{2}$

Hence, after reaction, volume of remaining $S O_{2}$

$= 1.6 - 1.4 L$

$= 0.2 L$

Q. At STP, 2.8 litres of hydrogen sulphide were mixed with 1.6 litres of sulphur dioxide and the reaction occurred according to the equation 2H2​S(g)+SO2​(g)→2H2​O(l)+3S(s) Which of the following shows the volume of the gas remaining after the reaction?

0.2 litres of SO2​(g)

0.4 litres of H2​(g)

1.2 litres of H2​S(g)

1.2 litres of SO2​(g)