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Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 2.71 g of ethane is

DanielleElmas [232]

Answer:

6g

Explanation:

Step 1:

The balanced equation for the reaction between gaseous ethane and gaseous oxygen. This is given below:

2C2H6 + 7O2 —> 4CO2 + 6H2O

Step 2:

Determination of the masses of C2H6 and O2 that reacted and the mass of CO2 produced from the balanced equation. This is illustrated below:

2C2H6 + 7O2 —> 4CO2 + 6H2O

Molar Mass of C2H6 = (12x2) + (6x1) = 24 + 6 = 30g/mol

Mass of C2H6 from the balanced equation = 2 x 30 = 60g

Molar Mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 7 x 32 = 224g

Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol

Mass of CO2 from the balanced equation = 4 x 44 = 176g

From the balanced equation above,

60g of C2H6 reacted with 224g of O2 to produce 176g of CO2.

Step 3:

Determination of the limiting reactant.

We need to determine the limiting reactant as it will be needed to obtain the maximum mass of CO2.

From the balanced equation above,

60g of C2H6 reacted with 224g of O2.

Therefore, 2.71g of C2H6 will react with = (2.71 x 224)/60 = 10.12g of O2.

From the above calculation, we can see that a higher mass of O2 is needed to react with 2.71g of C2H6, therefore, O2 is the limiting reactant.

Step 4:

Determination of the maximum mass of CO2 produced when 2.71 g of ethane is mixed with 7.6 g of oxygen.

The limiting reactant is used to determine the maximum mass.

From the balanced equation above,

224g of O2 produce 176g of CO2.

Therefore, 7.6g of O2 will produce = (7.6 x 176)/224 = 5.97g ≈ 6g of CO2

From the calculations made above, the maximum mass of CO2 produced is 5.97 ≈ 6g

5 0

3 years ago

What volume (mL) of the sweetened tea described in Example 1 contains the same amount of sugar (mol) as 10 mL of the soft drink

S_A_V [24]

The question is incomplete, the complete question is:

What volume (mL) of the sweetened tea described in Example 3.14 contains the same amount of sugar (mol) as 10 mL of the soft drink in this example. The example is attached below.

Answer: 75 mL of sweetened tea will contain the same amount of sugar as in 10 mL of soft drink

Explanation:

We first calculate the number of moles of soft drink in a volume of 10 mL

The formula used to calculate molarity:

$\text{Molarity of solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (mL)}}$ .....(1)

Taking the concentration of soft drink from the example be = 0.375 M

Volume of solution = 10 mL

Putting values in equation 1, we get:

$0.375=\frac{\text{Moles of sugar in soft drink}\times 1000}{10}\\\\\text{Moles of sugar in soft drink}=\frac{0.375\times 10}{1000}=0.00375mol$

Calculating volume of sweetened tea:

Moles of sugar = 0.00375 mol

Molarity of sweetened tea = 0.05 M

Putting values in equation 1, we get:

$0.05=\frac{0.00375\times 1000}{\text{Volume of sweetened tea}}\\\\\text{Volume of sweetened tea}=\frac{0.00375\times 1000}{0.05}=75mL$

Hence, 75 mL of sweetened tea will contain the same amount of sugar as in 10 mL of soft drink

5 0

3 years ago

you are investigating an organic compound. you discover the following: it contains no hydrogen atoms. what property do you expec

pashok25 [27]

This can be, for example, halogensubstituted hydrocarbons.
CCl₄, C₂F₆.

Or halides halocarboxylic acids, and other compounds.
O
II
Cl₃C-Cl

6 0

3 years ago

Read 2 more answers

How do solve for #13?What is the boiling point of a solution made by dissolving 1.0000 mole of sucrose in 1.0000 kg of water?

exis [7]

What is the boiling point of a solution made by dissolving 1.0000 mole of sucrose in 1.0000 kg of water?

The change in Boiling Point of water can be calculated using this formula:

ΔTb = i * Kb * m

Where i is the van't hoff factor (the number of particles or ions), the kb is a constant (boiling point elevation constant) and m is the molality of the solution.

The kb for water is always 0.515 °C/m. Kb = 0.515 °C/m

The value for i in this case is 1. Since sucrose is a covalent compound and it doesn't dissociate into ions. i = 1

The molal concentration of the solution can be found using this formula:

molality = moles of sucrose/kg of water

molality = 1.000 mol / 1.000 kg of water

molality = 1 m

Now that we know all the values, we can use the formula to find the change in the boiling point of water:

ΔTb = i * Kb * m

ΔTb = 1 * 0.515 °C/m * 1 m

ΔTb = 0.515 °C

Finally, we are asked for the boiling point of the solution, not the change. The boiling point of water at atmospheric pressure is 100.00 °C. If the boiling point rises 0.515 °C when we prepare the solution. The boiling point of the solution is:

Boiling point solution = Boiling point of water + ΔTb

Boiling point solution = 100.000 °C + 0.515 °C

Boiling point solution = 100.515 °C

Answer: The boiling point of the solution is 100.515 °C.

8 0

1 year ago

Hydrogen gas is filled in a 224 ml glass container at 0 degree celcius and 1 atmospheric pressure. What are the number of molecu

Step2247 [10]

Answer:

$\boxed{\text{6.02 $\math{\times 10^{21}}$ molecules}}$

Explanation:

A pressure of 1 atm and a temperature of 0 °C is the old definition of STP. Under these conditions, 1 mol of a gas occupies 22.4 L.

1. Calculate the moles of hydrogen.

$n = \text{0.224 L} \times \dfrac{\text{1 mol}}{\text{22.4 L}} = \text{0.0100 mol}$

2. Calculate the number of molecules

$\text{No. of molecules} = \text{0.0100 mol} \times \dfrac{\text{6.022 $\times 10^{23}$ molecules}}{\text{1 mol}}\\\\= \textbf{6.02 $\mathbf{\times 10^{21}}$ molecules}\\\\\text{The sample contains }\boxed{\textbf{6.02 $\mathbf{\times 10^{21}}$ molecules}}$

8 0

3 years ago

Read 2 more answers