What amount of oxygen is used at S.T.P. to obtain 9 g water from sufficient amount of hydrogen gas?
Show 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: $\frac{11.2}{2} = 5.6\ \text{dm}^3$!
Step 1: Write the reaction.
Water forms by $H_2 + \frac{1}{2}O_2 \rightarrow H_2O$. So one mole of water, that is 18 g, needs half a mole of oxygen.
Step 2: Volume of oxygen for 18 g water.
Half a mole of gas at STP is $0.5 \times 22.4 = 11.2$ dm$^3$.
Step 3: Scale to 9 g.
9 g is half of 18 g, so the oxygen volume is also halved: $\dfrac{11.2}{2} = 5.6$ dm$^3$.
\[ \boxed{5.6\ \text{dm}^3,\ \text{option 1}} \]