Question

What amount of oxygen is used at S.T.P. to obtain 9 g water from sufficient amount of hydrogen gas?

• A. $5.6\ \text{dm}^3$
• B. $22.4\ \text{dm}^3$
• C. $16.8\ \text{dm}^3$
• D. $11.2\ \text{dm}^3$

Hint:

Notice that 9 grams is exactly half of water's molar mass (18 grams). Since you are making half a mole of water, you only need half the usual volume of oxygen gas: $\frac{11.2}{2} = 5.6\ \text{dm}^3$!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem asks for the volume of oxygen gas consumed at Standard Temperature and Pressure (STP) to generate exactly 9 g of water vapor/liquid via a stoichiometric reaction with hydrogen gas.

Step 2: Key Formula or Approach:
First, we formulate the balanced chemical equation for the synthesis of water: $$ \text{H}_{2(g)} + \frac{1}{2}\text{O}_{2(g)} \longrightarrow \text{H}_2\text{O}_{(l)} $$ According to Avogadro's law, 1 mole of any ideal gas occupies a molar volume of $22.4\ \text{dm}^3$ at STP. Therefore, $\frac{1}{2}$ mole of $\text{O}_2$ gas corresponds to a volume of $11.2\ \text{dm}^3$.

Step 3: Detailed Explanation:
From our balanced chemical pathway:

• 1 mole of $\text{H}_2\text{O}$ has a mass equal to its molecular weight = $18\ \text{g}$.

• Producing 1 mole ($18\ \text{g}$) of $\text{H}_2\text{O}$ consumes exactly $\frac{1}{2}$ mole of $\text{O}_2$ gas.

• Volume of $\frac{1}{2}$ mole of $\text{O}_2$ at STP = $0.5 \times 22.4\ \text{dm}^3 = 11.2\ \text{dm}^3$.
Now, apply a simple unitary approach for our target mass of $9\ \text{g}$: $$ 18\ \text{g of water requires} \longrightarrow 11.2\ \text{dm}^3\ \text{of }\text{O}_2 $$ $$ 9\ \text{g of water requires} \longrightarrow \frac{11.2}{18} \times 9 = \frac{11.2}{2} = 5.6\ \text{dm}^3\ \text{of }\text{O}_2 $$

Step 4: Final Answer:
The volume of oxygen gas required at STP is $5.6\ \text{dm}^3$, corresponding to option (A).