To a container containing 3 moles of C2H6 further 60 gram C2H6 is added and 2.4×1024 molecules of gas are then removed. The left over gas is burnt in the presence of excess oxygen. Then :

nC2H6=3+6030−24×10236×1023=1 mole; C2H6+72O2→2CO2+3H2O12mol 3mol=30g=45.4L =54g

To a container containing 3 moles of C2H6 further 60 gram C2H6 is added and 2.4×1024 molecules of gas are then removed. The left over gas is burnt in the presence of excess oxygen. Then :

nC2H6=3+6030−24×10236×1023=1 mole; C2H6+72O2→2CO2+3H2O12mol 3mol=30g=45.4L =54g

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To a container containing 3 moles of $C_{2} H_{6}$ further 60 gram $C_{2} H_{6}$ is added and $2.4 \times 10^{24}$ molecules of gas are then removed. The left over gas is burnt in the presence of excess oxygen. Then :

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mass of liquid water produced is 54 gram

$30 gms$ of $C_{2} H_{6}$ are left for combustion

60gms of $C_{2} H_{6}$ are left for combustion

volume of ${CO}_{2}$ at STP produced is 45.4 litre

answer is B, C, D.

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Detailed Solution

$\begin{aligned} n_{C_{2} H_{6}} = 3 + \frac{60}{30} - \frac{24 \times 10^{23}}{6 \times 10^{23}} = 1 mole; \\ \begin{array}{ll} C_{2} H_{6} + \frac{7}{2} O_{2} & \to 2 {CO}_{2} + 3 H_{2} O \\ 1 & 2 mol 3 mol \\ = 30 g & = 45.4 L = 54 g \end{array} \end{aligned}$