If H2SO4 is formed from its elements by taking 6.023×1023 an atom of O,5.6 liter of H2 gas at 1 atm and 273 K and 8gm S, then:

OHS↓↓↓11214141414Nothing is left and 14mole H2SO4 is formed.Hence, options B and C are correct.

If H2SO4 is formed from its elements by taking 6.023×1023 an atom of O,5.6 liter of H2 gas at 1 atm and 273 K and 8gm S, then:

OHS↓↓↓11214141414Nothing is left and 14mole H2SO4 is formed.Hence, options B and C are correct.

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If $H_{2} S O_{4}$ is formed from its elements by taking $6.023 \times 10^{23}$ an atom of $O, 5.6$ liter of $H_{2}$ gas at $1 atm$ and $273 K$ and $8 gm S$ , then:

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$0.125$ moles of $H_{2} S O_{4}$ are formed

$0.25$ moles of $H_{2} S O_{4}$ are formed

no moles of $S$ are left

$\frac{1}{4}$ mole of $O_{2}$ is left

answer is B, C.

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$\begin{array}{ccc} O & H & S \\ ↓ & ↓ & ↓ \\ 1 & \frac{1}{2} & \frac{1}{4} \\ \frac{1}{4} & \frac{1}{4} & \frac{1}{4} \end{array}$

Nothing is left and $\frac{1}{4}$ mole $H_{2} S O_{4}$ is formed.

Hence, options B and C are correct.

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If H2SO4 is formed from its elements by taking 6.023×1023 an atom of O,5.6 liter of H2 gas at 1 atm and 273 K and 8gm S, then:

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If $H_{2} S O_{4}$ is formed from its elements by taking $6.023 \times 10^{23}$ an atom of $O, 5.6$ liter of $H_{2}$ gas at $1 atm$ and $273 K$ and $8 gm S$ , then: