Consider the balanced equation for the following reaction:7O2(g) + 2C2H6(g) → 4CO2(g) + 6H2O(l)
Determine the amount of CO2(g) formed in the reaction if 8.00 grams of O2(g) reacts with an excess of C2H6(g) and the percent yield of CO2(g) is 90.0%.

Answers

Answer 1
Answer:

Answer: The amount of carbon dioxide formed in the reaction is 5.663 grams

Explanation:

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

Given mass of oxygen gas = 8 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

\text{Moles of oxygen gas}=(8g)/(32g/mol)=0.25mol

For the given chemical equation:

7O_2(g)+2C_2H_6(g)\rightarrow 4CO_2(g)+6H_2O(l)

By Stoichiometry of the reaction:

7 moles of oxygen gas produces 4 moles of carbon dioxide

So, 0.25 moles of oxygen gas will produce = (4)/(7)* 0.25=0.143mol of carbon dioxide

Now, calculating the mass of carbon dioxide from equation 1, we get:

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = 0.143 moles

Putting values in equation 1, we get:

0.143mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\n\n\text{Mass of carbon dioxide}=(0.143mol* 44g/mol)=6.292g

To calculate the experimental yield of carbon dioxide, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}* 100

Percentage yield of carbon dioxide = 90 %

Theoretical yield of carbon dioxide = 6.292 g

Putting values in above equation, we get:

90=\frac{\text{Experimental yield of carbon dioxide}}{6.292g}* 100\n\n\text{Experimental yield of carbon dioxide}=(90* 6.292)/(100)=5.663g

Hence, the amount of carbon dioxide formed in the reaction is 5.663 grams


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Write the equilibrium expression, calculate KEQ and then tell where the equilibrium lies: Fe (s) + O2 (g) ↔ Fe2O3 (s) In a 2.0 L Container At equilibrium: Fe = 1.0 mol O2 = 1.0 E-3 mol Fe2O3 = 2.0 mol

Answers

Answer:

The value of Keq is 4e-9. See the solution below

Explanation:

We need to balanced rhe equation and use the formula of the Keq

A ballon has a volume of 5.6L at 25C and it is placed into liquid nitrogen at a temp of -78c. what is the volume of the ballon

Answers

Answer:

3.7L

Explanation:

Given parameters:

Initial volume  = 5.6L

Initial temperature  = 25°C = 273 + 25 = 298K

Final temperature = -78°C = 273 + (-78) = 195K

Unknown:

New volume of the balloon = ?

Solution:

We are going to assume that the pressure is constant and then we apply the charles's law;

  "the volume of fixed mass of a gas varies directly as its absolute temperature if its pressure is constant"

 mathematically;

         (V_(1) )/(T_(1) )   = (V_(2) )/(T_(2) )

Where V and T are temperature

            1 and 2 are initial and final states

Insert the parameters and solve;

       (5.6)/(298)  = (V_(2) )/(195)

      298V₂  = 1092

              V₂  = 3.7L

For the combustion reaction of C9H12 in O2: how many moles of O2 is required to react with 0.67 mol C9H12?

Answers

Answer:

8.0 mol O₂

Explanation:

Let's consider the complete combustion reaction of C₉H₁₂.

C₉H₁₂ + 12 O₂ → 9 CO₂ + 6 H₂O

The molar ratio of C₉H₁₂ to O₂ is 1:12. The moles of O₂ required to react with 0.67 moles of C₉H₁₂ are:

0.67 mol C₉H₁₂ × (12 mol O₂/1 mol C₉H₁₂) = 8.0 mol O₂

8.0 moles of O₂ are required to completely react with 0.67 moles of C₉H₁₂.

Answer:

To react with 0.67 moles C9H12 we need 8.04 moles of O2

Explanation:

Step 1: Data given

Number of moles C9H12 = 0.67 moles

Step 2: The balanced equation

C9H12 + 12O2 → 9CO2 + 6H2O

Step 3: Calculate moles of O2 required

For 1 mol C9H12 we need 12 moles of O2 to produce 9 moles of CO2 and 6 moles of H2O

For 0.67 moles of C9H12 we need 12 *0.67 = 8.04 moles of O2

To produce 9*0.67 = 6.03 moles of CO2 and 6*0.67 = 4.02 moles H2O

To react with 0.67 moles C9H12 we need 8.04 moles of O2

Please help me i’m in need!

Answers

Answer:

balanced............

On another planet, the isotopes of titanium have the following natural abundances. a. Isotope 46Ti Abundance 70.900% Mass(amu) 45.95263
b. Isotope 48Ti Abundance 10.000% Mass(amu) 47.94795
c. Isotope 50Ti Abundance 19.100% Mass(amu) 49.94479
d. What is the average atomic mass of titanium on that planet?
e. I got 46.9 amu but it is wrong.

Answers

Answer:

Average atomic mass = 46.91466 amu

Explanation:

Step 1: Data given

Isotopes of titanium

46Ti = 70.900% ⇒ 45.95263 amu

48Ti = 10.000 % ⇒ 47.94795 amu

50Ti = 19.100 % ⇒ 49.94479 amu

Step 2: Calculate the average atomic mass of titanium

Average atomic mass = 0.7090 * 45.95263 + 0.10 * 47.94795 + 0.1910 * 49.94479

Average atomic mass = 46.91466 amu

Answer True or False for each of the following statements. (a) The carburization surface was maintained at slightly less than 0.25 wt% carbon for each specimen. (b) Comparing the finished specimens at a depth of 0.20 mm, specimen A features the lowest carbon concentration. (c) Comparing the finished specimens as a whole, specimen D features the lowest overall amount of carbon.

Answers

Answer:

verdadero/a

falso/b

verdadero/c

Explanation: