All categories

3 years ago

2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(g)

Chemistry

1 answer:

Ket [755]3 years ago

5 0

Answer:- 10 L of ethane.

Solution:- The given balanced equation is:

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

From this equation, ethane and oxygen react in 2:7 mol ratio, the ratio of volumes would also be same if they are at same temperature and pressure.

Since 14 L of each gas are taken, the oxygen will be the limiting reactant and ethane will be the excess reactant. Let's calculate the volume of ethane used:

$14LO_2(\frac{2LC_2H_6}{7LO_2})$

= $4LC_2H_6$

From above calculations, 4 L of ethane are used. So, excess volume of ethane left after the completion of reaction = 14 L - 4 L = 10 L

Hence, 10 L of ethane will be remaining.

You might be interested in

Consider the resonance structures of formate. the first lewis structure of formate has a central carbon atom. a hydrogen atom an

OleMash [197]

So,

Formate has a resonating double bond.

In molecular orbital theory, the resonating electrons are actually delocalized and are shared between the two oxygens. So the carbon-oxygen bonds can be described as 1.5-bonds (option B). I'm not sure if option C is correct, however, because the likelihood of both delocalized electrons being in the area of one oxygen atom is less than 50%.<span />

6 0

3 years ago

1. How many grams of B are present in 3.35 grams of boron tribromide ?

Andre45 [30]

Answer:

The answer to your question is:

Explanation:

1. How many grams of B are present in 3.35 grams of boron tribromide ?

________ grams B.

MW BBr₃ = 251 g

251 g of BBr₃ ---------------------- 11 g of B

3.35 g ----------------------- x

x = (3.35 x 11) / 251 = 0.147 g of B

2. How many grams of boron tribromide contain 4.69 grams of Br ?

________grams boron tribromide.

MW BBr₃ = 251g

251g of BBr₃ ----------------- 80 g of Br

x ----------------- 4.69 g

x = (4.69 x 251)/ 80 = 14.71 g of BBr₃

3. How many grams of N are present in 4.11 grams of nitrogen trifluoride ?

________grams N.

MW NF₃ = 71 g

71 g of NF₃ ----------------- 14 g of N

4.11 g ---------------- x

x = (4.11 x 14) / 71 = 0.81 g of N

4. How many grams of nitrogen trifluoride contain 3.07 grams of F ?

________grams nitrogen trifluoride.

MW NF₃ = 71

71 g of NF₃ --------------------- 19 g of F

x --------------------- 3.07 g

x = (3.07 x 71) / 19 = 11.5 g of NF₃

5.How many grams of Co3+ are present in 1.16 grams of cobalt(III) iodide?

________grams Co3+.

MW CoI₃ = 440 g

440 g of CoI₃ ------------------ 59 g of Co

1.16 g ------------------ x

x = (1.16 x 59) / 440 = 0.16 g of Co

6. How many grams of cobalt(III) iodide contain 2.28 grams of Co3+?

________grams cobalt(III) iodide.

MW CoI₃ = 440 g

440 g of CoI₃ ------------------ 59 g of Co

x ----------------- 2.28 g of Co⁺³

x = (2.28 x 440) / 59

x = 17 g of CoI₃

5 0

3 years ago

Why is the atomic weight of carbon reported as 12.011 in the table of elements and the periodic table in the front inside cover

Oduvanchick [21]

The atomic weight reported in the periodic table of elements for Carbon is the average weight of all the known isotopes of Carbon occurring naturally. There are 3: Carbon-12, Carbon-13 and Carbon-14. The numbers represent the mass numbers, or the sum of neutrons and protons in the nucleus. Each of the individual masses are multiplied with their individual percentage abundances. Then, when you add them up, it will yield the average atomic weight equal to 12.011 amu or grams/mole C.

5 0

3 years ago

Ne, c4h10, is a component of natural gas that is used as fuel for cigarette lighters. the balanced equation of the complete comb

Simora [160]

<span>3.68 liters First, determine the number of moles of butane you have. Start with the atomic weights of the involved elements: Atomic weight carbon = 12.0107 Atomic weight hydrogen = 1.00794 Atomic weight oxygen = 15.999 Molar mass butane = 4*12.0107 + 10*1.00794 = 58.1222 g/mol Moles butane = 2.20 g / 58.1222 g/mol = 0.037851286 Looking at the balanced equation for the reaction which is 2 C4H10(g)+13 O2(g)â†’8 CO2(g)+10 H2O(l) It indicates that for every 2 moles of butane used, 8 moles of carbon dioxide is produced. Simplified, for each mole of butane, 4 moles of CO2 are produced. So let's calculate how many moles of CO2 we have: 0.037851286 mol * 4 = 0.151405143 mol The ideal gas law is PV = nRT where P = Pressure V = Volume n = number of moles R = Ideal gas constant ( 0.082057338 L*atm/(K*mol) ) T = absolute temperature (23C + 273.15K = 296.15K) So let's solve the formula for V and the calculate using known values: PV = nRT V = nRT/P V = (0.151405143 mol) (0.082057338 L*atm/(K*mol))(296.15K)/(1 atm) V = (3.679338871 L*atm)/(1 atm) V = 3.679338871 L So the volume of CO2 produced will occupy 3.68 liters.</span>

4 0

3 years ago

What is plasama?.......

Andrej [43]

Explanation:

Plasma is the clear, straw-colored liquid portion of blood that remains after red blood cells, white blood cells, platelets and other cellular components are removed. It is the single largest component of human blood, comprising about 55 percent, and contains water, salts, enzymes, antibodies and other proteins.

5 0

3 years ago

Read 2 more answers