# Question #7da6e

Aug 2, 2017

#### Answer:

${H}_{2} + \frac{1}{2} {O}_{2} \to {H}_{2} O$

#### Explanation:

Atomic weights: 1 g (H) and 16 g (O).

You have excess oxygen since 16 grams of oxygen and 2 grams of hydrogen yield 18 grams of water. Therefore, your limiting reactant is oxygen.

If 16 grams of oxygen and 2 grams of hydrogen form 18 grams of ${H}_{2} O$

You wonder how much water is formed if you have 20 grams of oxygen and more than sufficient amount of hydrogen.

Your answer is

$= \frac{20 \times 18}{16} = 22.5$ grams of water.

You will have an excessive amount of hydrogen which is 1.5 grams.

Aug 2, 2017

#### Answer:

Well, mass is conserved, so we are going to end up with $24 \cdot g$ of oxygen and hydrogen, whatever the compound/elemental makeup.

#### Explanation:

We follow the stoichiometric equation......

${H}_{2} \left(g\right) + \frac{1}{2} {O}_{2} \left(g\right) \rightarrow {H}_{2} O \left(l\right)$

ANd so dihydrogen, and dioxygen react in a 2:1 molar relationship....

$\text{Moles of dihydrogen} = \frac{4 \cdot g}{2.0 \cdot g \cdot m o {l}^{-} 1} = 2 \cdot m o l$

$\text{Moles of dioxygen} = \frac{20 \cdot g}{32.0 \cdot g \cdot m o {l}^{-} 1} = 0.625 \cdot m o l$

Clearly dioxygen is the limiting reagent, the reagent in stoichiometric deficiency, and thus, given the reaction we should form.......

$0.625 \cdot m o l \times 2 \times 18.01 \cdot g \cdot m o {l}^{-} 1 = 22.5 \cdot g$ of water.

What is the mass of hydrogen left over?