Question

MULTI CORRECT
From 3 moles of C2H6 gas 45 g gas is removed, it is followed by the removal of 3 x 1023 molecules
further.
The leftover gas is combusted in the presence of excess oxygen then (NA = 6 x 1023)
(Density of H2O is 1g/mL).
(A) 2 moles of C2Ho left for combustion
(B) Volume of CO2 at STP produced after combustion 44.8 litres.
(C) Volume of water produced is 54 mL
(D) None​

Answer 1

Answer:

(b) and (c) is     the answer

Explanation:

molar mass of c2h6 is 30 g per mol

so we have 90g at the beginig.

next 45  g were removed    

so we now have 45 g left with us

next half moles were removed

so we are left with 30 g c2h6 ( half mole = 15 g) which is equal to 1 mole

now the reaction equation is c2h6 + 3.5o2 = 2co2 + 3h2o

so one mole c2h6 gives 2 mole co2 = 2*22.4L=44.8 L            (B)

and one mole c2h6 gives 3 mole water = 18*3 g (molar mass of water =18g per mol) = 54g= 54ml ( density of water is 1g per ml) (C)

Answer 2

answer : (B) and (C)

explanation : initially, no of moles of C2H6 = 3

now, given C2H6 is mixed with 60g of this gas.

so, no of mole of gas is mixed = 60g/30g = 2 [ as molar mass of C2H6 = 30g/mol]

so, total no of moles in mixture = 3 + 2 = 5 moles.

now, 2.4 × 10²⁴ molecules of gas is removed.

no of mole of 2.4 × 10²⁴ molecules = 2.4 × 10²⁴/6 × 10²³ = 4 moles.

so, no of moles of remaining C2H6 gas = 5moles - 4moles = 1 mol

now, 1mole of C2H6 is burnt in the presence of excess oxygen.

$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

here it is clear that, 1 mole of C2H6 produced 2 moles of CO2 and 3 moles of H2O.

so, volume of 2mol of CO2 at STP = 2 × 22.4 L = 44.8 L

hence, option (B) is correct.

volume of liquid water = (3mol × 18g/mol)/1g/ml = 54ml [ molar mass of water = 18g/mol]

hence, option (c) is also correct.