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2 years ago

2H2 + O2- 2H20

Chemistry

1 answer:

tankabanditka [31]2 years ago

7 0

Conversion factor : molar mass/molecular weight

Mass of H₂O produced = 72 g

<h3>Further explanation</h3>

Given

Reaction

2H2 + O2- 2H20

8 grams of H2

Required

Conversion factors

Solution

We can use the molar mass of the components and the mole ratio as conversion factors for the above equation

mol H₂ (MW = 2 g/mol) :

$\tt mol=\dfrac{8}{2}=4~mol$

From the equation, mol ratio of H₂ : H₂O = 2 : 2, so mol H₂O = 4 mol

mass H₂O(MW = 18 g/mol) :

mass = mol x MW

mass = 4 x 18

mass = 72 g

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A 3.00-kg block of copper at 23.0°C is dropped into a large vessel of liquid nitrogen at 77.3 K. How many kilograms of nitrogen

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1.2584kg of nitrogen boils.

Explanation:

Consider the energy balance for the overall process. There are not heat or work fluxes to the system, so the total energy keeps the same.

For the explanation, the 1 and 2 subscripts will mean initial and final state, and C and N2 superscripts will mean copper and nitrogen respectively; also, liq and vap will mean liquid and vapor phase respectively.

The overall energy balance for the whole system is:

$U_1=U_2$

The state 1 is just composed by two phases, the solid copper and the liquid nitrogen, so: $U_1=U_1^C+U_1^{N_2}$

The state 2 is, by the other hand, composed by three phases, solid copper, liquid nitrogen and vapor nitrogen, so:

$U_2=U_2^C+U_{2,liq}^{N_2}+U_{2,vap}^{N_2}$

So, the overall energy balance is:

$U_1^C+U_1^{N_2}=U_2^C+U_{2,liq}^{N_2}+U_{2,vap}^{N_2}$

Reorganizing,

$U_1^C-U_2^C=U_{2,liq}^{N_2}+U_{2,vap}^{N_2}-U_1^{N_2}$

The left part of the equation can be written in terms of the copper Cp because for solids and liquids Cp≅Cv. The right part of the equation is written in terms of masses and specific internal energy:

$m_C*Cp*(T_1^C-T_2^C)=m_{2,liq}^{N_2}u_{2,liq}^{N_2}+m_{2,vap}^{N_2}u_{2,vap}^{N_2}-m_1^{N_2}u_1^{N_2}$

Take in mind that, for the mass balance for nitrogen, $m_1^{N_2}=m_{2,liq}^{N_2}+m_{2,vap}^{N_2}$ ,

So, let's replace $m_1^{N_2}$ in the energy balance:

$m_C*Cp*(T_1^C-T_2^C)=m_{2,liq}^{N_2}u_{2,liq}^{N_2}+m_{2,vap}^{N_2}u_{2,vap}^{N_2}-m_{2,liq}^{N_2}u_1^{N_2}-m_{2,vap}^{N_2}u_1^{N_2}$

So, as you can see, the term $m_{2,liq}^{N_2}u_{2,liq}^{N_2}$ disappear because $u_{2,liq}^{N_2}=u_{1,liq}^{N_2}$ (The specific energy in the liquid is the same because the temperature does not change).

$m_C*Cp*(T_1^C-T_2^C)=m_{2,vap}^{N_2}u_{2,vap}^{N_2}-m_{2,vap}^{N_2}u_1^{N_2}$

$m_C*Cp*(T_1^C-T_2^C)=m_{2,vap}^{N_2}(u_{2,vap}^{N_2}-u_1^{N_2})$

The difference $(u_{2,vap}^{N_2}-u_1^{N_2})$ is the latent heat of vaporization because is the specific energy difference between the vapor and the liquid phases, so:

$m_{2,vap}^{N_2}=\frac{m_C*Cp*(T_1^C-T_2^C)}{(u_{2,vap}^{N_2}-u_1^{N_2})}$

$m_{2,vap}^{N_2}=\frac{3kg*0.092\frac{cal}{gC} *(296.15K-77.3K)}{48.0\frac{cal}{g}}\\m_{2,vap}^{N_2}=1.2584kg$

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3 years ago