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2 years ago

How many grams of aluminum in required to produce 17.5 g of hydrogen

Chemistry

1 answer:

azamat2 years ago

6 0

Mass of Aluminum = 157.5 g

<h3>Further explanation</h3>

Given

17.5 g of Hydrogen

Required

Mass of Aluminum

Solution

Reaction

2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂

mol Hydrogen :

= mass : molar mass

= 17.5 g : 2 g/mol

= 8.75

From the equation, mol Al :

= 2/3 x mol H₂

= 2/3 x 8.75

= 5.83

Mass Al :

= 5.83 x 27 g/mol

= 157.5 g

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Acetylene reacts with oxygen to form carbon dioxide and water according to the following unbalanced reaction:

Lelechka [254]

Answer:

Mass of excess reactant leftover after reaction is complete = 34.2 g $\mathrm{C}_{2} \mathrm{H}_{2}$ .

Explanation:

Acetylene reacts with oxygen to form carbon dioxide and water, equation given is unbalanced

$\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})$

Firstly equation is balanced:

$2 \mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 4 \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$

$\text { Moles of } \mathrm{CO}_{2} \text { from } \mathrm{C}_{2} \mathrm{H}_{2}$

$\text { Molar mass of } \mathrm{C}_{2} \mathrm{H}_{2} \text { is } 26.04 \mathrm{g} / \mathrm{mol}$

$\text { Moles of } \mathrm{C}_{2} \mathrm{H}_{2}=38.7 \mathrm{gm} \mathrm{C}_{2} \mathrm{H}_{2} \times \frac{1 \mathrm{mol} c_{2} \mathrm{H}_{2}}{26.04 \mathrm{gm} \mathrm{c}_{2} \mathrm{H}_{2}}=1.48 \mathrm{mol} \mathrm{C}_{2} \mathrm{H}_{2}$

$1.48 \mathrm{mol} \mathrm{C}_{2} \mathrm{H}_{2} \times \frac{4 \mathrm{mol} \mathrm{Co}_{2}}{2 \mathrm{mol} C_{2} \mathrm{H}_{2}}=2.96 \mathrm{mol} \mathrm{CO}_{2}$

$\text { Moles of } \mathrm{CO}_{2} \text { from } \mathrm{O}_{2}$ :

$\text { Molar mass of } \mathrm{O}_{2} \text { is } 32.00 \mathrm{g} / \mathrm{mol}$

$\text { Moles of } \mathrm{O}_{2}=13.7 \mathrm{gm} \mathrm{O}_{2} \times \frac{1 \mathrm{mol} O_{2}}{32.00 \mathrm{gm} 0_{2}}=0.428 \mathrm{mol} \mathrm{O}_{2}$

$\text { Here we come to know that } 4 \mathrm{mol} \mathrm{CO}_{2} \equiv 5 \mathrm{mol} \mathrm{O}_{2}, \text { hence }$

$0.428 \mathrm{mol} \mathrm{O}_{2} \times \frac{4 \mathrm{mol} \mathrm{co}_{2}}{5 \mathrm{mol} O_{3}}\left(4 \mathrm{mol} \mathrm{CO}_{2}\right) /\left(5 \mathrm{mol} \mathrm{O}_{2}\right)=0.3424 \mathrm{mol} \mathrm{CO}_{2}$

Here we come to know O2 is limiting reactant as it gives smaller amount of $\mathrm{CO}_{2}$ So $\mathrm{C}_{2} \mathrm{H}_{2}$ is excess reactant.

Mass of excess reactant utilized:

$0.428 \mathrm{mol} \mathrm{O}_{2} \times \frac{2 \mathrm{mol} C_{2} \mathrm{H}_{2}}{5 \mathrm{mol} O_{2}}=0.1712 \mathrm{mol} \mathrm{C}_{2} \mathrm{H}_{2}$

$0.1712 \mathrm{mol} \mathrm{C}_{2} \mathrm{H}_{2} \times \frac{26.04 \mathrm{gm} \mathrm{c}_{2} \mathrm{H}_{2}}{1 \mathrm{mol} C_{2} \mathrm{H}_{2}}=4.45 \mathrm{gm} \mathrm{C}_{2} \mathrm{H}_{2}$

Mass of excess reactant leftover after reaction is complete:

$38.7 \mathrm{gm} \mathrm{C}_{2} \mathrm{H}_{2}-4.45 \mathrm{gm} \mathrm{C}_{2} \mathrm{H}_{2}=34.2 \mathrm{gm} \mathrm{C}_{2} \mathrm{H}_{2}$ .

7 0

3 years ago

vekshin1

Typically, the actual yield is lower than the theoretical yield because few reactions complete (i.e., are not 100% efficient) or because not all of the product in a reaction is collected. It is also conceivable for the real yield to exceed the theoretical yield.

What is theoretical yield ?

Theoretical yield is what would be obtained if 100% of the reaction was completed and no product was wasted in any way; however, procedures are not this efficient. Everything is dependent on the experimenter's precision and the reaction. Some reactions are reversible, meaning that less than 100% of the reaction proceeds to completion; some reactions do not go to completion because they require a large amount of energy or additional time, for example. Substances in the reaction may be lost throughout the method or difficult to separate from other sections of the experiment. In chemistry, some degree of inaccuracy is always expected.

To learn more about theoretical yield
brainly.com/question/25996347
#SPJ1

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1 year ago

2. Rank the boiling points of methane, methanol, and methanal from lowest to highest. Explain your ranking. (6 points) 3. The lo

zlopas [31]

Answer:

(2) Boiling point order: Methanol (highest) > Methanal > Methane (weakest)

(3) Boiling point of alcohol will be higher than ester molecules

Explanation:

(2) Methane is a non-polar molecule. Hence only weakest van der waal inter molecular force is present between methane molecules.

Methanal is polar molecule due to presence of polar aldehyde group. hence weaker dipole-dipole inter molecular force is present between methanal molecules.

Methanol is a polar protic molecule. Hence strongest H-bonding force act between methanol molecules.

The stronger the inter molecular force, the higher will be boiling point.

Boiling point order: Methanol (highest) > Methanal > Methane (weakest)

(3) An alcohol is a polar protic molecule. Hence strongest H-bonding force exist between alcohol molecules.

An ester is polar molecule. Therefore weaker dipole-dipole inter molecular force is present between ester molecules.

So boiling point of alcohol will be higher than ester molecules.

6 0

3 years ago

When potassium carbonate and calcium chromate are mixed, which of the following will pass through a filter?

aliina [53]

Calcium carbonate will be formed which is insoluble in water.

The answer is a) K and CrO4 only.

8 0

3 years ago

Name CH3CH2CH2CH=CHCH3

blondinia [14]

Answer:

Hexene

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3 years ago

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