Ask question

Ask question

All categories

2 years ago

12

A mixture of 0.10 mol of NO, 0.050 mol of H2, and 0.10 mol of H2O is placed in a 1.0-L vessel at 300 K. The following equilibriu

m is established: 2NO(g)+2H2(g)⥫⥬==N2(g)+2H2O(g) At equilibrium [NO]=0.062M. Calculate the equilibrium concentrations of H2, N2, and H2O.
Calculate the equilibrium concentration of N2

1 answer:

tatyana61 [14]2 years ago

5 0

Answer:

[H2] = 0.012 M

[N2] = 0.019 M

[H2O] = 0.057 M

Explanation:

The strategy here is to account for the species at equilibrium given that the concentration of [NO]=0.062M at equilibrium is known and the quantities initially present and its stoichiometry.

2NO(g) + 2H2(g) ⇒ N2(g) + 2H2O(g)

i mol 0.10 0.050 0.10

c mol -0.038 -0.038 +0019 +0.038

e mol 0.062 0.012 00.019 0.057

Since the volume of the vessel is 1.0 L, the concentrations in molarity are:

[NO] = 0.062 M

[H2] = 0.012 M

[N2] = 0.019 M

[H2O] = 0.057 M

You might be interested in

What is the difference between melting point and boiling point?

MrRissso [65]

The melting point is the temperature a substance must reach to go into its liquid phase (or solid) and a boiling point is the temperature a substance must reach to go into its gas phase (or liquid). For example, water must be 0ºC to melt or freeze and 100ºC to boil.

6 0

2 years ago

Read 2 more answers

Ce cantitate de O se gaseste in 5,8 g hidroxid de magneziu

coldgirl [10]

Answer:

separare pâlnii de picurare balon cotat. 1. 2. 3. 4 5 6 7. 8. 9. 1 ... Mod de lucru: 25 g Na2SO4∙7H2O se dizolvă în cantitatea minim ... Exemple: NaOH – hidroxid de sodiu.

Explanation:

make me as brain liest

4 0

3 years ago

A 25.0 mL aliquot of 0.0680 M EDTA was added to a 59.0 mL solution containing an unknown concentration of V3 . All of the V3 pre

givi [52]

Answer:

$\mathbf{0.02 M}$

Explanation:

$\text{So, from the given question:}$

$\text{EDTA will make complex with}$ $V^{+3}$ $\text{and the remaining EDTA will react with }$ $Ga^{+3}$

$\text{Hence, the total concentration of}$ $V^{+3}$ & $Ga^{+3}$ $\text{will be equivalent to EDTA concentration.}$

$V_{EDTA} = 25 \ mL$

$V_{V^{+3}} = 59.0 \ mL$

$V_{Ga^{+3}} = 13.0 \ mL$

$M_{EDTA} = 0.0680 \ M$

$M_{V^{+3}} = ???(unknown)$

$M_{Ga^{+3}} = 0.0400 \ M$

$V^{+3} + EDTA \to V[EDTA] + EDTA(Excess) \to^{CoA} \ Ga[EDTA] _{complex}$

$M_{EDTA} \times V_{EDTA} = ( V_{V^+3}\times M_{V^{+3}}+ V_{Ga^{+3} }\times M_{Ga^{+3}}})$

$0.0680 \times 25 = (59\times x + 13 \times 0.040) \\ \\ 1.7 = 59x + 0.52\\ \\ 1.7 - 0.52 = 59x \\ \\ 59x = 1.18$

$x = \dfrac{1.18}{59}$

$\mathbf{x =0.02 \ M }$

5 0

3 years ago

1. Burning coal and oil in a power plant produces pollutants, such as sulfur dioxide, SO2. This compound can be removed by the f

Lorico [155]

Idk tbh srry man am in quarantine so I have a big packet to do

3 0

3 years ago

Using the molarity equation, calculate how many grams of salt you need in order to make 80ml of a 2M solution (MW = 58.44g/mole)

ollegr [7]

Answer:

9.35g

Explanation:

The molarity equation establishes that:

$\textrm{molarity}=\frac{\textrm{moles of solute}}{\textrm{liters of solution}}$

So, we have information about molarity (2M) and volume (80 ml=0.08 l), with that, we can find the moles of solute:

$\textrm{moles of solute}=\textrm{molarity}*\textrm{liters of solution}$

$\textrm{moles of solute}= 0.08 \textrm{ l} *2\textrm{ M} = 0.16 \textrm{ mol}$

The mathematical equation that establishes the relationship between molar weight, mass and moles is:

$\textrm{molar weight}= \frac{\textrm{mass}}{\textrm{moles}}$

$\textrm{MW}= \frac{\textrm{m}}{\textrm{n}}$

We have MW (58.44g/mole) and n (0.16 mol), and we need to find m (grams of salt needed) to solve the problem:

$\textrm{m} = \textrm{MW * n}= 58.44\frac{\textrm{g}}{\textrm{mol}} * 0.16 \textrm{ mol} = 9.35 \textrm{ g}$

8 0

3 years ago