Question

Fluorine gas and water vapor react to form hydrogen fluoride gas and oxygen. What volume of hydrogen fluoride would be produced by this reaction if of water were consumed? Also, be sure your answer has a unit symbol, and is rounded to the correct number of significant digits.

Answer 1

Answer:

Hi, the question is incomplete. However, the question is about the calculation of volume of a product when the volume of one of the reactants is provided.

9.587 cm^3

Explanation:

The balanced equation for the chemical reaction is shown below:

$2F_{2(g)} + 2H_{2}O_{(g)}$ ⇒$4HF_{(g)} + O_{2(g)}$

In the chemical reaction above, 2 moles of water produced 4 moles of hydrogen fluoride. If 4.8 cm^3 of water were consumed, we can calculated the volume of hydrogen fluoride that would be produced as follow:

Using STP, 1 mole of gas has a volume of 22.4 L

Thus, 4.8 cm^3 = 0.0048 L is equivalent to 2.14*10^-4

since 2 moles of water produced 4 moles of hydrogen fluoride, therefore, 2.14*10^-4 would produced 2*2.14*10^-4 = 4.28*10^-4 moles

we can convert the moles to L by multiplying with 22.4

volume of hydrogen fluoride = 4.28*10^-4 * 22.4 = 0.009587 L = 9.587 cm^3

Answer 2

I believe the question is incomplete and assume the complete question is this: Fluorine gas and water vapor react to form hydrogen fluoride gas and oxygen. what volume of oxygen would be produced by this reaction if 7.3 l of water were consumed? also, be sure your answer has a unit symbol, and is rounded to 2 significant digits.

Answer:

3.7L of oxygen will be produced by 7.3L of water vapor.

Explanation:

First, write an equation for the above reaction:

F2(g) + H20 (g) → HF(g) + 02(g)

Next, balance the equation:

2F2(g) + 2H2O(g) → 4HF(g) + O2(g)

From the balanced equation,

2 mols of water vapor produced 1mol of oxygen

But, 1 mol of any gas at Standard Temperature and Pressure ( S.T.P. ) occupies a volume = 22.4L

This implies that,

2 x 22.4L (44.8L) of water vapor produced 22.4L of oxygen,

So, 7.3L of water vapor will produce,

7.3L x 22.4L/44.8L = 163.52/44.8 = 3.65L of oxygen = 3.7L of oxygen ( to 2 significant digits ).