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3 years ago

What is the mass of 2.15 liters of N2 gas at STP?

Chemistry

1 answer:

devlian [24]3 years ago

6 0

Answer:

2.68g

Explanation:

Given parameters:

Volume of gas at STP = 2.15L

Unkown:

Mass of Nitrogen N₂ gas at STP = ?

Solution

To find the mass of the gas at STP, we use the mole concept approach.

In using the mole approach we follow the following procedures:

1. Find the number of moles of gas at STP using the relationship below:

Number of moles = $\frac{volume occupied}{22.4dm^{3} mol^{-1} }$

Note: 1L = 1dm³

2. Then using the mole and mass relationship, we can find the mass of the gas using the equation below:

Mass = number of moles x molar mass

Workings

1. Number of moles of N₂ = $\frac{2.15}{22.4}$

Number of moles of N₂ = 0.096mole

2. Given that the atomic mass of N = 14g

Molar mass of N₂ = 2 x 14 = 28gmol⁻¹

Mass = 0.096mol x 28gmol⁻¹ = 2.68g

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he thermite reaction, in which powdered aluminum reacts with iron(III) oxide, is highly exothermic. 2 Al(s) + Fe2O3(s) Al2O3(s)

Afina-wow [57]

The given question is incomplete. The complete question is

The thermite reaction, in which powdered aluminum reacts with iron oxide, is highly exothermic: $2Al(s)+Fe_2O_3(s)\rightarrow Al_2O_3(s)+2Fe(s)$ . Use standard enthalpies of formation to find ΔH∘rxn for the thermite reaction. Express the heat of the reaction in kilojoules to four significant figures.

Answer: ΔH∘rxn for the thermite reaction is -851.5 kJ

Explanation:

The balanced chemical reaction is :

$2Al(s)+Fe_2O_3(s)\rightarrow Al_2O_3(s)+2Fe(s)$

We have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{Fe}\times \Delta H_f^0_{(Fe)}+n_{Al_2O_3}\times \Delta H_f^0_{(Al_2O_3)}]-[n_{Al}\times \Delta H_f^0_(Al)+n_{Fe_2O_3}\times \Delta H_f^0_{(Fe_2O_3)}]$

where,

We are given:

$\Delta H^o_f_{(Fe_2O_3(s))}=-824.2kJ/mol\\\Delta H^o_f_{(Al_2O_3(s))}=-1675.7kJ/mol\\\Delta H^o_f_{(Fe(s))}=0kJ/mol\\\Delta H^o_f_{(Al(s)}=0kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(2\times 0)+(1\times -1675.5)]-[(2\times 0)+(1\times -824.2)]=-851.5kJ$

ΔH∘rxn for the thermite reaction is -851.5 kJ

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3 years ago

A gas mixture has a total pressure of 0.51 atm and consists of He and Ne. If the partial pressure of He in the mixture is 0.32 a

sp2606 [1]

Explanation:

Relation between total pressure and mole fraction is as follows.

$P_{1} = P^{o}x_{1}$

where, $P_{1}$ = partial pressure of gas 1

$x_{1}$ = mole fraction of gas 1

$P^{o}$ = total pressure of the gases

Also, it is known that $x_{1} + x_{2}$ = 1

or, $x_{2} = 1 - x_{1}$

It is given that total pressure of gas mixture is 0.51 atm and partial pressure of helium is 0.32 atm.

Hence, calculate the mole fraction of helium gas present in the given mixture as follows.

$P_{He} = P^{o}x_{He}$

0.32 atm = $0.51 atm \times x_{He}$

$x_{He}$ = 0.63

As, $x_{He} + x_{Ne}$ = 1

so, $x_{Ne} = 1 - x_{He}$

= 1 - 0.63

= 0.37

Hence, calculate the partial pressure of Ne as follows.

$P_{total} = P_{He}x_{He} + P_{Ne}x_{Ne}$

$0.51 atm = 0.32 atm \times 0.63 + P_{Ne}\times 0.37$

$0.51 atm = 0.2016 atm + 0.37 \times P_{Ne}$

0.3084 atm = $0.37 \times P_{Ne}$

$P_{Ne} = \frac{0.3084 atm}{0.37}$

= 0.834 atm

Thus, we can conclude that the partial pressure of the Ne in the mixture is 0.834 atm.

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