All categories

3 years ago

How many Liters of I2 are produced when 241.6 grams of KI are reacted with excess CuBr2?

Chemistry

1 answer:

Viefleur [7K]3 years ago

3 0

Answer:

8.15 L I2

Explanation:

<u>So, I have the balanced equation of</u>: 2CuBr2 +4KI → 2CuI + 4KBr + I2

<u>Now, we do</u>:

241.6g / 1 mol KI / 1 mol I2 / 22.4 L

--------------------------------------------------------- = 8.15 L I2

/ 166g KI / 4 mol KI / 1 mol I2

Hope this helps! ^u^

You might be interested in

Gravitational Potential Energy is created when an object’s what changes

alisha [4.7K]

Answer:

option b is the correct answer

5 0

3 years ago

What total volume of ozone measured at a pressure of 24.5 mmHg and a temperature of 232 K can be destroyed when all of the chlor

ddd [48]

Answer:

$1.75272\ \text{m}^3$

Explanation:

The breakdown reaction of ozone is as follows

$CF_3Cl + UV \rightarrow CF_3 + Cl$

$Cl + O_3 \rightarrow ClO + O_2$

$O_3 + UV \rightarrow O_2 + O$

$ClO + O \rightarrow Cl + O_2$

It can be seen that 2 moles of ozone is required in the complete cycle

So for 10 cycles, 20 moles of ozone is required

m = Mass of $CF_3Cl$ = 15.5 g

M = Molar mass of $CF_3Cl$ = 104.46 g/mol

P = Pressure = 24.5 mmHg

T = Temperature = 232 K

R = Gas constant = $62.363\ \text{L mmHg/K mol}$

Number of moles is given by

$n=\dfrac{m}{M}\\\Rightarrow n=\dfrac{15.5}{104.46}\\\Rightarrow n=0.1484\ \text{moles}$

$20\ \text{moles} = 20\times 0.1484 = 2.968\ \text{moles}$

From ideal gas law we have

$PV=nRT\\\Rightarrow V=\dfrac{nRT}{P}\\\Rightarrow V=\dfrac{2.968\times 62.363\times 232}{24.5}\\\Rightarrow V=1752.72\ \text{L}=1.75272\ \text{m}^3$

For 20 cycles of the reaction the volume of the ozone is $1.75272\ \text{m}^3$ .

7 0

3 years ago

nswer the following questions relating to HCl, CH3Cl, and CH3Br.HCl(g), can be prepared by the reaction of concentrated H2SO4(ag

alexdok [17]

Answer:

It is an example of double displacement reaction.

4.8 g of NaCl is needed to react.

Explanation:

Balanced reaction: $H_{2}SO_{4}(aq.)+2NaCl(s)\rightarrow 2HCl(g)+Na_{2}SO_{4}(aq.)$

Here, oxidation states of H, S, O, Na and Cl do not change during reaction. Hence it is not a redox reaction.

In this reaction, cations and anions of the reactants interchange their partners during reaction. Hence, it is an example of double displacement reaction.

As $H_{2}SO_{4}(aq.)$ remain in excess amount therefore NaCl (s) is the limiting reagent. Hence production of HCl entirely depends on amount of NaCl used.

Molar mass of HCl = 36.46 g/mol

So, 3.0 g of HCl = $\frac{3.0}{36.46}$ mol of HCl = 0.082 mol of HCl

According to balanced equation-

2 moles of HCl are produced from 2 moles of NaCl

So, 0.082 moles of HCl are produced from 0.082 moles of NaCl

Molar mass of NaCl = 58.44 g/mol

So, mass of 0.082 moles of NaCl = $(0.082\times 58.44)$ g = 4.8 g

Hence 4.8 g of NaCl is needed to react.

4 0

3 years ago

g In Part 7, the [Cl-] in saturated NaCl is 5.4 M at room temperature. Assume that you had 1.00 ml of the saturatedsolution, and

trapecia [35]

Answer:

7.60 M

Explanation:

Our method to solve this question is to use the definition of molarity (M) concentration which is the number of moles per liter of solution, so for this problem we have

[Cl⁻] = # mol Cl⁻ / Vol

Now the number of moles of Cl⁻ will be sum of Cl in the 1.00 mL 5.4 M solution plus the moles of Cl⁻ in the 0.50 mL 12 M H . Since the volume in liters times the molarity gives us the number of moles we will have previous conversion of volume to liters for units consistency:

1mL x 1 L / 1000 mL = 0.001 L

0.5 mL x 1L/1000 mL = 0.0005 L

[Cl⁻] = 0.001 L x 5.4 mol/L + 0.0005L x 12 mol/L / ( 0.001 L + 00005 L )

= 7.6 M

This is the same as the statement given in the question.

5 0

3 years ago

Nurse Antonio measured out 7 grams of sodium chloride (NaCI). Using dimensional analysis, calculate how many moles of NaCI he we

Neporo4naja [7]

Answer:

0.119 moles

Explanation:

Given that,

Mass measured by Nurse Antonio is 7 grams of NaCl

To find,

The no of moles of NaCl

Solution,

The number of moles is given by

$n=\dfrac{m}{M}$

m is given mass

M is molar mass

For NaCl, molar mass is 23+35.5 = 58.5 grams

So,

$n=\dfrac{7}{58.5}\\\\n=0.119\ \text{moles}$

Therefore, there are 0.119 moles of NaCl.

6 0

3 years ago