Question

If 1.6 g of SO2 naf 1.5× 10^22 molecules of H2S arr mixed and allowed to remain in contact in a closed vessel until the reaction. 2H2S + SO2 gives 3S + 2H2O. (1) only S and H2O remain in the reaction vassel. (2) H2S will reamain in excess.(3) SO2 will reamain in excess (4) None .​

Answer 1

Answer:

option 3

Explanation:

from the equation given,2 moles of H2S require 1 mole of SO2

=1 mole of H2S requires 0.5 moles of SO2

=34 g of H2S requres 1/2*64=32 i.e 34g of  h2s requires 32g of so2( using atomic mass)

∴  1g of h2s requires 32/34g of so2

∴1.6g of h2s requres32/34*1.6 g so2 which gives 1.50g⇒ eq no 1

now find the no of moles of so2 and then find the weight  and you will get it as 1.6g

as so2 is present in more than sufficient quantity i.e 1.6g (when the required quantity is 1.5g (from eq no 1))

therefore option 3 -so2 will remain in excess

Answer 2

Answer: (3) $SO_2$ will remain in excess.

Explanation:

To calculate the moles, we use the equation:

1.$\text{Number of moles}=\frac{\text{Given mass}}{\textMolar mass}}$

$\text{Number of moles} of SO_2=\frac{1.6g}{64g/mol}=0.025moles$

2. $\text{Number of moles}=\frac{\text{Given molecules}}{\text {Avogadro's no}}$

$\text{Number of moles} of H_2S=\frac{1.5\times 10^{22}}{6.023\times 10^{23}}=0.025moles$

$2H_2S+SO_2\rightarrow 3S+2H_2O$

By stoichiometry of the reaction:

2 moles of $H_2S$ reacts with 1 mole of $SO_2$

Thus 0.025 moles of $H_2S$ will react with =$\frac{1}{2}\times 0.025=0.012moles$ of $SO_2$

Thus $H_2S$ will act as limiting reagent as it limits the formation of product and $SO_2$ acts as excess reagent and (0.025-0.012)=0.012 moles of $SO_2$ are left as such.