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:
[tex]2F_{2(g)} + 2H_{2}O_{(g)}[/tex] ⇒[tex]4HF_{(g)} + O_{2(g)}[/tex]
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
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 ).
