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Amiraneli [1.4K]
3 years ago
10

Using the Hoffman apparatus for electrolysis, a chemist decomposes 2.3 moles of water into its gaseous elements. How many grams

of hydrogen gas should get (theoretical yield)? The chemist collected 2.0 moles hydrogen gas. What is his percent yield?
Chemistry
1 answer:
solong [7]3 years ago
7 0

Answer:

2.318032g

Explanation:

-The electrolysis equation of water is written as below:

2H_2O_{(l)}->2H_2_{(g)}+O_2_{(g)}

-The mole ratio of Water to the hydrogen formed is 1:1, therefore 2.3 moles of hydrogen gas is produced.

-Hydrogen's molar mass is 1.00784 grams:

Mass=moles\times molar \ mass\\\\\=2.3\times 1.00784\\\\=2.318032\ g

Hence, 2.318032 grams of hydrogen is produced.

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3. If you start with 8x1025 molecules of Cl, and 25 grams of KI, how many grams of KCl would
miskamm [114]

Answer:

Percent yield = 89.1%

Explanation:

Based on the equation:

Cl₂ + 2KI → 2KCl + I₂

<em>1 mole of Cl₂ reacts with 2 moles of KI to produce to moles of KCl</em>

<em />

To solve this quesiton we must find the moles of each reactant in order to find the limiting reactant. With the limiting reactant we can find the moles of KCl and the mass:

<em>Moles Cl₂:</em>

8x10²⁵ molecules * (1mol / 6.022x10²³ molecules) = 133 moles

<em>Moles KI -Molar mass: 166.0028g/mol-</em>

25g * (1mol / 166.0028g) = 0.15 moles

Here, clarely, the KI is the limiting reactant

As 2 moles of KI produce 2 moles of KCl, the moles of KCl produced are 0.15 moles. The theoretical mass is:

0.15 moles * (74.5513g / mol) =

11.2g KCl

Percent yield is: Actual yield (10.0g) / Theoretical yield (11.2g) * 100

<h3>Percent yield = 89.1%</h3>
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