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anyanavicka [17]
2 years ago
11

A sample of gas contains 0.1700 mol of OF2(g) and 0.1700 mol of H2O(g) and occupies a volume of 19.5 L. The following reaction t

akes place: OF2(g) + H2O(g)O2(g) + 2HF(g) Calculate the volume of the sample after the reaction takes place, assuming that the temperature and the pressure remain constant. L
Chemistry
1 answer:
andreev551 [17]2 years ago
8 0

Answer: The volume of the sample after the reaction takes place is 29.25 L.

Explanation:

The given reaction equation is as follows.

OF_{2}(g) + H_{2}O(g) \rightarrow O_{2}(g) + 2HF(g)

So, moles of product formed are calculated as follows.

\frac{3}{2} \times 0.17 mol \\= 0.255 mol

Hence, the given data is as follows.

n_{1} = 0.17 mol,      n_{2} = 0.255 mol

V_{1} = 19.5 L,         V_{2} = ?

As the temperature and pressure are constant. Hence, formula used to calculate the volume of sample after the reaction is as follows.

\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}

Substitute the values into above formula as follows.

\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}\\\frac{19.5 L}{0.17 mol} = \frac{V_{2}}{0.255 mol}\\V_{2} = \frac{19.5 L \times 0.255 mol}{0.17 mol}\\= \frac{4.9725}{0.17} L\\= 29.25 L

Thus, we can conclude that the volume of the sample after the reaction takes place is 29.25 L.

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A 3.3 g sample of sodium hydrogen carbonate is added to a solution of acetic acid weighing 10.3 g. The two substances react, rel
Zanzabum

Answer:

1.73g of CO2.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

NaHCO3 + CH3COOH → CH3COONa + H2O + CO2

Next we shall determine the masses of NaHCO3 and CH3COOH that reacted and the mass of CO2 produced from the balanced equation. This is illustrated below:

Molar mass of NaHCO3 = 23 + 1 + 12 + (16x3) = 84g/mol

Mass of NaHCO3 from the balanced equation = 1 x 84 = 84g

Molar mass of CH3COOH = 12 + (3x1) + 12 + 16 + 16 + 1 = 60g/mol

Mass of CH3COOH from the balanced equation = 1 x 60 = 60g

Molar mass of CO3 = 12 + (2x16) = 44g/mol

Mass of CO2 from the balanced equation = 1 x 44 = 44g

From the balanced equation above,

84g of NaHCO3 reacted with 60g of CH3COOH to produce 44g of CO2.

Next, we shall determine the limiting reactant of the reaction. This is illustrated below:

From the balanced equation above,

84g of NaHCO3 reacted with 60g of CH3COOH.

Therefore, 3.3g of NaHCO3 will react with = (3.3 x 60)/84 = 2.36g of CH3COOH.

From the above illustration, we can see that only 2.36g of CH3COOH out of 10.3g given reacted completely with 3.3g of NaHCO3. Therefore, NaHCO3 is the limiting reactant while CH3COOH is the excess reactant.

Finally, can determine the mass of CO2 produced during the reaction.

In this case the limiting reactant will be used because it will produce the mass yield of CO2 as all of it were used up in the reaction. The limiting reactant is NaHCO3 and the mass of CO2 produced is obtained as shown below:

From the balanced equation above,

84g of NaHCO3 reacted to produce 44g of CO2.

Therefore, 3.3g of NaHCO3 will react to produce = (3.3 x 44)/84 = 1.73g of CO2.

Therefore, 1.73g of CO2 is released during the reaction.

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