Question

Hydrogen chloride gas and oxygen react to form water vapor and chlorine gas. What volume of water would be produced by this reaction if 6.5 L of hydrogen chloride were consumed? Also, be sure your answer has a unit symbol, and is rounded to the correct number of significant digits.

Answer 1

When 6.5 liters of hydrogen chloride gas react, 3.3 liters of water vapor will be produced, following the stoichiometry of the balanced chemical equation and Avogadro's Law.

Step-by-step explanation:

The student is asking about a chemical reaction between hydrogen chloride gas and oxygen to produce water vapor and chlorine gas. To calculate the volume of water vapor produced, we need the balanced chemical equation for the reaction:

4 HCl(g) + O₂(g) → 2 H₂O(g) + 2 Cl₂(g)

According to the stoichiometry of the balanced equation, four moles of HCl react with one mole of O₂ to produce two moles of H₂O and two moles of Cl₂. Given 6.5 L of HCl gas consumed in this reaction, assuming all reactants are at the same conditions of temperature and pressure, we can use Avogadro's Law (equal volumes of gases at the same temperature and pressure contain the same number of molecules) to determine the volume of H₂O produced.

From the stoichiometry, we can see that four volumes of HCl produce two volumes of H₂O. Therefore, we divide the volume of HCl by two to find the volume of water vapor produced:

Volume of water vapor = (6.5 L) / 2 = 3.25 L

We need to round this to two significant digits because 6.5 has two significant digits, so the final answer is:

Volume of water vapor produced = 3.3 L

Answer 2

Answer: 3.4 L

Step-by-step explanation:

To calculate the moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given Volume}}{\text {Molar Volume}}=(6.5L)/(22.4L)=0.29moles`

The balanced chemical equation for reaction of Hydrogen chloride gas with oxygen to form water vapor and chlorine gas.

`4HCl(g)+O_2(g)\rightarrow 2H_2O(g)+2Cl_2(g)`

According to stoichiometry :

4 moles of
`HCl` produce = 2 moles of
`H_2O`

Thus 0.29 moles of
`HCl` will produce =
`(2)/(4)* 0.29=0.15moles` of
`H_2O`

Volume of
`H_2O=moles* {\text {Molar Volume}}=0.15moles* 22.4L/mol=3.4L`

Thus 3.4 L of of water would be produced by this reaction if 6.5 L of hydrogen chloride were consumed