Ask question

Ask question

All categories

2 years ago

12

A gas phase mixture of H2 and N2 has a total pressure of 784 torr with an H2 partial pressure of 124 torr. What mass of N2 gas i

s present in 2.00 L of the mixture at 298 K

1 answer:

olchik [2.2K]2 years ago

4 0

The mass of N₂ in the mixture having a total pressure of 784 torr with an H₂ partial pressure of 124 torr is 1.988 g

We'll begin by calculating the partial pressure of N₂. This can be obtained as follow:

Total pressure = 784 torr

Partial pressure of H₂ = 124 torr

<h3>Partial pressure of N₂ =?</h3>

Total pressure = Partial pressure of H₂ + Partial pressure of N₂

784 = 124 + Partial pressure of N₂

Collect like terms

Partial pressure of N₂ = 784 – 124

<h3>Partial pressure of N₂ = 660 Torr</h3>

Next, we shall determine the number of mole of N₂ in the mixture.

Pressure of N₂ (P) = 660 Torr

Volume (V) = 2 L

Temperature (T) = 298 K

Gas constant (R) = 62.364 L.Torr/Kmol

<h3>Number of mole of N₂ (n) =? </h3>

PV = nRT

660 × 2 = n × 62.364 × 298

1320 = n × 18584.472

Divide both side by 18584.472

n = 1320 / 18584.472

<h3>n = 0.071 mole</h3>

Finally, we shall determine the mass of N₂ in the mixture.

Mole of N₂ = 0.071 mole

Molar mass of N₂ = 2 × 14 = 28 g/mol

<h3>Mass of N₂ =? </h3>

Mass = mole × molar mass

Mass of N₂ = 0.071 × 28

<h3>Mass of N₂ = 1.988 g</h3>

Therefore, the mass of N₂ in the mixture having a total pressure of 784 torr with an H₂ partial pressure of 124 torr is 1.988 g

Learn more: brainly.com/question/20853110

You might be interested in

Consider this reaction: 6 CO2 + 6 H2O + light equation C6H12O6 + 6 O2 If there were 2.38 x 102 g of H2O, 18.6 moles of CO2, and

alisha [4.7K]

H₂O would be the limiting reactant.

Balanced chemical equation:

6CO₂ + 6H₂O + light equation → C₆H₁₂O₆ + 6O₂

The amount of product that can be created is constrained by the reactant that is consumed first in a chemical reaction, commonly referred to as the limiting reactant (or limiting reagent).

Given

No. of moles of CO₂ = 18.6

Mass of H₂O = 2.38 × 10² g = 238g

No. of moles of H₂O = Given mass/ Molar mass

= 238 / 18 = 13.22 moles

Moles of H₂O = 13.22

According to the balanced chemical equation

6 moles of CO₂ react with 6 moles of H₂O

So the reactant that has less number of moles will be consumed first.

As the No. of moles of H₂O < No. of moles of CO₂

So, H₂O is the limiting reactant with 13.22 moles.

Hence, H₂O would be the limiting reactant.

Learn more about limiting reactant here brainly.com/question/14222359

#SPJ1

7 0

2 years ago

True or false? Steel is a chemical element.

Irina-Kira [14]

I think answer is false

8 0

3 years ago

Read 2 more answers

Phosgene (COCl2) is a toxic substance that forms readily from carbon monoxide and chlorine at elevated temperatures: CO(g) + Cl2

lapo4ka [179]

Answer: Concentration of $CO$ = 0.328 M

Concentration of $Cl_2$ = 0.328 M

Concentration of $COCl_2$ = 0.532 M

Explanation:

Moles of $CO$ and $Cl_2$ = 0.430 mole

Volume of solution = 0.500 L

Initial concentration of $CO$ and $Cl_2$ = $\frac{moles}{volume}=\frac{0.430}{0.500}=0.860M$

The given balanced equilibrium reaction is,

$CO(g)+Cl_2(g)\rightleftharpoons COCl_2(g)$

Initial conc. 0.860M 0.860M 0

At eqm. conc. (0.860-x) M (0.860-x) M (x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[COCl_2]}{[CO][Cl_2]}$

Now put all the given values in this expression, we get :

$4.95=\frac{x}{(0.860-x)^2}$

By solving the term 'x', we get :

x = 0.532 M

Thus, the concentrations of $CO,Cl_2\text{ and }COCl_2$ at equilibrium are :

Concentration of $CO$ = (0.860-x) M =(0.860-0.532) M = 0.328 M

Concentration of $Cl_2$ = (0.860-x) M = (0.860-0.532) M = 0.328 M

Concentration of $COCl_2$ = x M = 0.532 M

3 0

3 years ago

How can we tell when something is polar or non-polar in an Inter-molecular Force?

Montano1993 [528]

If the substance mixes with water it's polar. If it doesn't it ms non polar.

7 0

3 years ago

In a 0.730 M solution, a weak acid is 12.5% dissociated. Calculate Ka of the acid.

Mamont248 [21]

Answer:

Approximately $1.30 \times 10^{-2}$ , assuming that this acid is monoprotic.

Explanation:

Assume that this acid is monoprotic. Let $\rm HA$ denote this acid.

$\rm HA \rightleftharpoons H^{+} + A^{-}$ .

Initial concentration of $\rm HA$ without any dissociation:

$[{\rm HA}] = 0.730\; \rm mol \cdot L^{-1}$ .

After $12.5\%$ of that was dissociated, the concentration of both $\rm H^{+}$ and $\rm A^{-}$ (conjugate base of this acid) would become:

$12.5\% \times 0.730\; \rm mol \cdot L^{-1} = 0.09125\; \rm mol \cdot L^{-1}$ .

Concentration of $\rm HA$ in the solution after dissociation:

$(1 - 12.5\%) \times 0.730\; \rm mol \cdot L^{-1} = 0.63875\; \rm mol\cdot L^{-1}$ .

Let $[{\rm HA}]$ , $[{\rm H}^{+}]$ , and $[{\rm A}^{-}]$ denote the concentration (in $\rm mol \cdot L^{-1}$ or $\rm M$ ) of the corresponding species at equilibrium. Calculate the acid dissociation constant $K_{\rm a}$ for $\rm HA$ , under the assumption that this acid is monoprotic:

$\begin{aligned}K_{\rm a} &= \frac{[{\rm H}^{+}] \cdot [{\rm A}^{-}]}{[{\rm HA}]} \\ &= \frac{(0.09125\; \rm mol \cdot L^{-1}) \times (0.09125\; \rm mol \cdot L^{-1})}{0.63875\; \rm mol \cdot L^{-1}}\\[0.5em]&\approx 1.30 \times 10^{-2} \end{aligned}$ .

5 0

3 years ago