Which is a proper description of chemical equilibrium?

What mass of Fe(OH)3 is produced when 35 mL of 0.250 M Fe(NO3)3 solution is mixed with 55 mL of a 0.180 M

Zina [86]

Answer:

0.35 g.

Explanation:

We'll begin by calculating the number of mole of Fe(NO3)3 in 35 mL of 0.250 M Fe(NO3)3 solution.

This is illustrated below:

Molarity of Fe(NO3)3 = 0.250 M

Volume = 35 mL = 35/1000 = 0.035 L

Mole of Fe(NO3)3 =?

Molarity = mole /Volume

0.250 = mole of Fe(NO3)3 / 0.035

Cross multiply

Mole of Fe(NO3)3 = 0.25 x 0.035

Mole of Fe(NO3)3 = 8.75×10¯³ mole

Next, we shall determine the number of mole of KOH in 55 mL of 0.180 M

KOH solution. This is illustrated below:

Molarity of KOH = 0.180 M

Volume = 55 mL = 55/1000 = 0.055 L

Mole of KOH =.?

Molarity = mole /Volume

0.180 = mole of KOH /0.055

Cross multiply

Mole of KOH = 0.180 x 0.055

Mole of KOH = 9.9×10¯³ mole.

Next, we shall write the balanced equation for the reaction. This is given below:

3KOH + Fe(NO3)3 —> Fe(OH)3 + 3KNO3

From the balanced equation above,

3 moles of KOH reacted with 1 mole of Fe(NO3)3 to produce 1 mole of Fe(OH)3.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

3 moles of KOH reacted with 1 mole of Fe(NO3)3.

Therefore, 9.9×10¯³ mole of KOH will react with = (9.9×10¯³ x 1)/3 = 3.3×10¯³ mole of Fe(NO3)3.

From the above illustration, we can see that only 3.3×10¯³ mole out of 8.75×10¯³ mole of Fe(NO3)3 given is needed to react completely with 9.9×10¯³ mole of KOH.

Therefore, KOH is the limiting reactant and Fe(NO3)3 is the excess reactant.

Next, we shall determine the number of mole of Fe(OH)3 produced from the reaction.

In this case, we shall use the limiting reactant because it will give the maximum yield of Fe(OH)3 as all of it is consumed in the reaction.

The limiting reactant is KOH and the mole of Fe(OH)3 produce can be obtained as follow:

From the balanced equation above,

3 moles of KOH reacted to produce 1 mole of Fe(OH)3.

Therefore, 9.9×10¯³ mole of KOH will react to produce = (9.9×10¯³ x 1)/3 = 3.3×10¯³ mole of Fe(OH)3.

Finally, we shall convert 3.3×10¯³ mole of Fe(OH)3 to grams. This can be obtained as follow:

Molar mass of Fe(OH)3 = 56 + 3(16 + 1) = 56 + 3(17) = 107 g/mol

Mole of Fe(OH)3 = 3.3×10¯³ mole

Mass of Fe(OH)3 =?

Mole = mass /Molar mass

3.3×10¯³ = Mass of Fe(OH)3 / 107

Cross multiply

Mass of Fe(OH)3 = 3.3×10¯³ x 107

Mass of Fe(OH)3 = 0.3531 ≈ 0.35 g.

Therefore, 0.35 g of Fe(OH)3 was produced from the reaction.

8 0

3 years ago

Why are space probes and other unmanned crafts more commonly used for space investigations than space crafts with crews of astro

Crank

All of these are correct.

4 0

3 years ago

How many molecules of carbon dioxide are in 9.080 x 10 ^-1 mol?

kakasveta [241]

Answer:

5.46 8 x 10²³ molecules.

Explanation:

Knowing that every one mole of a substance contains Avogadro's no. of molecules (NA = 6.022 x 10²³).

Using cross multiplication:

1.0 mole → 6.022 x 10²³ molecules.

9.08 x 10⁻¹ mole → ??? molecules.

∴ The no. of molecules of CO₂ are in 9.08 x 10⁻¹ mol = (6.022 x 10²³ molecules) ( 9.08 x 10⁻¹ mole) / (1.0 mol) = 5.46 8 x 10²³ molecules.

6 0

3 years ago

A gas occupies 200ml at a temperature of 26 degrees Celsius and 76mmHg pressure. Find the volume at -3degree Celsius with the pr

sergey [27]

Answer:

184.62 ml

Explanation:

Let $p_1, v_1,$ and $T_1$ be the initial and $p_2, v_2,$ and $T_2$ be the final pressure, volume, and temperature of the gas respectively.

Given that the pressure remains constant, so

$p_1=p_2$ ...(i)

$v_1$ = 200 ml

$T_1= 26 ^{\circ}C = 273+26 =299$ K

$T_2= 3 ^{\circ}C = 273+3 =276$ K

From the ideal gas equation, pv=mRT

Where p is the pressure, v is the volume, T is the temperature in Kelvin, m is the mass of air in kg, R is the specific gas constant.

For the initial condition,

$p_1v_1=mRT_1 \\\\mR= \frac{p_1v_1}{T_1}\cdots(ii)$

For the final condition,

$p_2v_2=mRT_2 \\\\mR= \frac{p_2v_2}{T_2}\cdots(iii)$

Equating equation (i), and (ii)

$\frac{p_1v_1}{T_1}=\frac{p_2v_2}{T_2}$

$\frac{v_1}{T_1}=\frac{v_2}{T_2}$ [from equation (i)]

$v_2=\frac{T_2}{T_1} \times v_1$

Putting all the given values, we have

$v_2=\frac{276}{299} \times 200 = 184.62 \; ml$

Hence, the volume of the gas at 3 degrees Celsius is 184.62 ml.

7 0

3 years ago

Ion If we add an electron to a stable element we call it a(n)_____ion anion cation

masha68 [24]

The answer is a cation, I may be wrong but cation is plus and anion is subtract

3 0

3 years ago