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%.

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:

.....(1)

Given mass of oxygen gas = 8 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

For the given chemical equation:

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 = 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:

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

Percentage yield of carbon dioxide = 90 %

Theoretical yield of carbon dioxide = 6.292 g

Putting values in above equation, we get:

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

Related Questions

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

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

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;

=

Where V and T are temperature

1 and 2 are initial and final states

Insert the parameters and solve;

=

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?

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.

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

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.

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.