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3 years ago

Stoichiometry

1 answer:

8 0

Stoichiometry is the ratio of reactants to products in a balanced chemical reaction equation.
the stoichiometry in this case of H₂ to O₂ is 2:1.
This means that 2 moles of H₂ reacts with 1 mol of O₂.
we have been asked to calculate the volume of H₂ gas.
Since both O₂ and H₂ are gases, at standard conditions its stated that molar volume of gases is 22.4 L
This means that at standard temperature and pressure, 1 mol of any gas occupies a volume of 22.4 L
first we need to calculate the number of O₂ moles reacted;
22.4 L of gas - 1 mol of O₂
1 L of gas - 1/22.4 mol/L
then 20 L of O₂ - 1/22.4 mol/L * 20 L = 0.89 mol
stoichiometry of H₂ to O₂ is 2:1
the number of H₂ moles = 0.89*2 = 1.79 mol
1 mol occupies 22.4 L
Therefore 1.79 mol = 22.4 L/mol * 1.79 mol = 40 L
Therefore it can be seen that stoichiometry applies to volumes as well.
volume of H₂ : O₂ = 40 L : 20 L = 2:1
volume and moles both can be determined by stoichiometry.
Volume of H₂ reacted = 40 L

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Answer:

B. accepted value x 0.1

Explanation:

in the equation provided

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Maximum allowed value of percentage error = 10%

put this value in the equation in stead of percentage error we get,

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My name is Ann [436]

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umka2103 [35]

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Ethane, C2H6, burns in oxygen. First write a balanced equation for this combustion reaction. What mass of oxygen, in grams, is r

iren [92.7K]

Answer:

We need 50.6 grams of oxygen

Explanation:

Step 1: Data given

Mass of ethane = 13.6 grams

Molar mass of ethane = 30.07 g/mol

Step 2: The balanced equation

2C2H6 + 7O2 → 4CO2 + 6H2O

Step 3: Calculate moles ethane

Moles ethane = mass ethane / molar mass ethane

Moles ethane = 13.6 grams / 30.07 g/mol

Moles ethane = 0.452 moles

Step 4: Calculate moles oxygen

For 2 moles ethane we need 7 moles O2 to produce 4 moles CO2 and 6 moles H2O

For 0.452 moles ethane we need 3.5*0.452 = 1.582 moles O2

Step 5: Calculate mass O2

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Let's assume you were given 2.0 g benzil, 2.2 g dibenzyl ketone, 50 mL 95% ethanol and 0.3 g potassium hydroxide to synthesize t

almond37 [142]

Answer:

the % yield is 82%

Explanation:

Given the data in the question,

we know that;

Molar mass of benzil is 210.23 g·mol−1

Molar mass of dibenzyl ketone is 210.27 g·mol−1

Molar mass of tetraphenylcyclopentadienone is 384.5 g·mol−1

Now,

2.0 g benzil = 2 g / 210.23 g·mol−1 = 0.0095 mole

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Now, the limiting reagent is benzil. 0.0095 mole can reacts wiyh 0.0095 mole of dibenzyl ketone

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= 0.82 × 100%

= 82%

Therefore, the % yield is 82%

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