Question

if 1 litre of O2 is formed from 2 moles of H2O2 ,what is the mass of H2O2 is consumed to form 2H2O+O2 when T=25°C,and P=1atm​

Answer 1

Answer:

•w=−PΔV=−Δn

w=−PΔV=−Δn g

w=−PΔV=−Δn g

w=−PΔV=−Δn g RT=−

w=−PΔV=−Δn g RT=− 34×2

w=−PΔV=−Δn g RT=− 34×24.5

w=−PΔV=−Δn g RT=− 34×24.5

w=−PΔV=−Δn g RT=− 34×24.5 ×8.314×298

w=−PΔV=−Δn g RT=− 34×24.5 ×8.314×298w=−1.63×10

w=−PΔV=−Δn g RT=− 34×24.5 ×8.314×298w=−1.63×10 2

w=−PΔV=−Δn g RT=− 34×24.5 ×8.314×298w=−1.63×10 2 J

Answer 2

Given:

The  1Litre of $O_{2}$ is formed from 2 moles of $H_{2} O_{2}$,$2H_{2}O+O_{2}$ when $T=25^{0}C$,and $P=1atm$​.

To Find:

The mass of $H_{2}O_{2}$.

Explanation:

The molality rate of $H_{2}O_{2}$ is $M_{r} =34.01$

       $2H_{2}O_{2}$→$2H_{2}O+O_{2}$

Convert grams of $H_{2}O_{2}$ to Moles $H_{2}O_{2}$

   $MolesofH_{2}O_{2}=1H_{2}O_{2}+\frac{1moleH_{2}O_{2}}{34.01 H_{2}O_{2}} \\\\=0.029molH_{2}O_{2}$

Convert moles of $H_{2}O_{2}$ to moles of $O_{2}$

$Moles of O_{2}=0.029MolH_{2}O_{2}*\frac{1 molO_{2}}{2molH_{2}O_{2}} \\\\=0.0143molO_{2}$

Calculate the volume of $O_{2}$

$pV=nRT$

Rearrange the formula

$V=\frac{nRT}{P}$

$n=0.0145mol\\R=0.08206LatmK^{-1}mol^{-1}\\T=(28+273.15)K=301.15K\\p=1atm$

$v=\frac{0.0145mol*0.082LatmK^{-1}mol^{-1}*301.15K}{1atm}\\=0.35L$

Answer:

Therefore, The mass of $H_{2}O_{2}$ is 0.35L .