Question

masscalc1. determine how many grams of oxygen are required to react with 4.0 g of hydrogen according to the following reaction.
$2\\ \ce{h_{2}}\\ \\ +\\ \\ \ce{o_{2}}\\ \\ \
ightarrow\\ \\ 2\\ \ce{h_{2}o}$
molar masses : $\ce{h_{2}}$ : $2.02\\ \mathrm{g/mol}$ $\ce{o_{2}}$ : $32.($
$\bigcirc$ 64 g
$\bigcirc$ 16 g
$\bigcirc$ 8.0 g
$\bigcirc$ 32 g

Explanation:

Step1: Calculate moles of \(H_2\)

Moles of \(H_2 = \frac{\text{mass of } H_2}{\text{molar mass of } H_2} = \frac{4.0\ g}{2.02\ g/mol} \approx 1.98\ mol\) (approx 2 mol for simplicity)

Step2: Use stoichiometry from reaction

From \(2H_2 + O_2
ightarrow 2H_2O\), 2 moles of \(H_2\) react with 1 mole of \(O_2\). So moles of \(O_2 = \frac{1}{2} \times\) moles of \(H_2\). For 2 mol \(H_2\), moles of \(O_2 = 1\) mol.

Step3: Calculate mass of \(O_2\)

Mass of \(O_2 = \text{moles of } O_2 \times \text{molar mass of } O_2 = 1\ mol \times 32\ g/mol = 32\ g\) (Note: Using approximate moles of \(H_2\) as 2 mol gives exact ratio here. If we use 1.98 mol \(H_2\), moles of \(O_2 \approx 0.99\) mol, mass \(\approx 0.99 \times 32 \approx 31.68\ g \approx 32\ g\))

Answer:

32 g (corresponding to the option "32 g")