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4 years ago

Using 18.8L, calculate the volume (in L) of water vapour that should be produced by the reaction of 35.0 g

Chemistry

1 answer:

victus00 [196]4 years ago

4 0

Answer:

A. 18.8L

B. 75.2L of H2O.

Explanation:

A. Determination of the volume of 35g of C3H8.

Date obtained from the question include the following:

Mass of C3H8 = 35g

Temperature (T) = 40°C

Pressure (P) = 110KPa

Volume (V) =..?

Next, we shall determine the number of mole (n) in 35g of C3H8. This is illustrated below:

Molar mass of C3H8 = (3x12) + (8x1) = 44g/mol

Mass of C3H8 = 35g

Mole of C3H8 =..?

Mole = mass /molar mass

Mole of C3H8 = 35/44

Mole of C3H8 = 0.795 mole

Finally, we shall determine the volume of 35g of C3H8 as follow:

Temperature (T) = 40°C = 40°C + 273 = 313K

Pressure (P) = 110KPa

Number of mole (n) = 0.795 mole

Gas constant (R) = 8.314 KPa.L/Kmol

Volume (V) =..?

PV = nRT

110 x V = 0.795 x 8.314 x 313

Divide both side by 110

V = (0.795 x 8.314 x 313)/110

V = 18.8L

Therefore, the volume of 35g of C3H8 under the conditions given is 18.8L

B. Determination of the volume of water vapour produced by the reaction of 35g of propane, C3H8.

From the calculations made in (A) above, 35g of C3H8 is equivalent to 18.8L of C3H8.

Thus, we can obtain the volume of water vapour produced as follow:

C3H8 + 5O2 —> 3CO2 + 4H2O

From the balanced equation above,

1L of C3H8 reacted to produce 4L of H2O.

Therefore, 18.8L of C3H8 will react to produce = (18.8 x 4)/1 = 75.2L of H2O.

Therefore, 75.2L of H2O were produced from the reaction.

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Answer:

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3 years ago

Consider the balanced equation for the following reaction:

Zanzabum

<u>Answer:</u> The amount of carbon dioxide formed in the reaction is 5.663 grams

<u>Explanation:</u>

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:

$\text{Moles of oxygen gas}=\frac{8g}{32g/mol}=0.25mol$

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 = $\frac{4}{7}\times 0.25=0.143mol$ 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:

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

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

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Percentage yield of carbon dioxide = 90 %

Theoretical yield of carbon dioxide = 6.292 g

Putting values in above equation, we get:

$90=\frac{\text{Experimental yield of carbon dioxide}}{6.292g}\times 100\\\\\text{Experimental yield of carbon dioxide}=\frac{90\times 6.292}{100}=5.663g$

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