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1 year ago

4 b + 3 o2 → 2 b2o3 if 8 moles of b and 4 moles of o2 are allowed to react, how many moles of b2o3 can be formed?

Chemistry

1 answer:

raketka [301]1 year ago

6 0

The number of moles of b2o3 that will be formed is determined as 4 moles.

<h3>Limiting reagent</h3>

The limiting reagent is the reactant that will be completely used up.

4 b + 3O₂ → 2b₂O₃

from the equation above;

4 b ------------> 2 b₂O₃

2b ------------> b₂O₃

2 : 1

3O₂ -------------> 2b₂O₃

3 : 2

b is the limiting reagent, thus, the amount of b2o3 to be formed is calculated as;

4 b ------------> 2 moles of b2o3

8 moles -------> ?

= (8 x 2)/4

= 4 moles

Thus, the number of moles of b2o3 that will be formed is determined as 4 moles.

Learn more about limiting reactants here: brainly.com/question/14222359

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What is the mass of 2.70 ×10^22 molecules of NaOH (Molar mass = 40.0 g/mol)?

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Data:
Molar Mass of NaOH = 40 g/mol

Solving: According to the Law Avogradro, we have in 1 mole of a substance, 6.02x10²³ atoms/mol or molecules

1 mol -------------------- 6.02*10²³ molecules
y mol -------------------- 2.70*10²² molecules

6.02*10²³y = 0.270*10²³
$y = \frac{0.270*\diagup\!\!\!\!\!\!10^{\diagup\!\!\!\!\!\!23}}{6.02*\diagup\!\!\!\!\!\!10^{\diagup\!\!\!\!\!\!23}}$
$\boxed{y \approx 0.04\:mol}$

Solving: Find the mass value now

40 g ----------------- 1 mol of NaOH
x g ------------- 0.04 mol of NaOH

$x = 40*0.04$
$\boxed{\boxed{x = 1.6\:g}}\end{array}}\qquad\quad\checkmark$

Answer:
The mass is 1.6 grams

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3 years ago

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3 years ago

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Particals in the nucleus of an atom are its _________ and ____________.

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2 years ago

Valproic acid, used to treat seizures and bipolar disorder, is composed of C, H, and O. A 0.165-g sample is combusted to produce

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Answer:

The empirical formula is = $C_4H_8O$

The formula of Valproic acid = $C_8H_{16}O_2$

Explanation:

Mass of water obtained = 0.166 g

Molar mass of water = 18 g/mol

Moles of $H_2O$ = 0.166 g /18 g/mol = 0.00922 moles

2 moles of hydrogen atoms are present in 1 mole of water. So,

Moles of H = 2 x 0.00922 = 0.01844 moles

Molar mass of H atom = 1.008 g/mol

Mass of H in molecule = 0.01844 x 1.008 = 0.018588 g

Mass of carbon dioxide obtained = 0.403 g

Molar mass of carbon dioxide = 44.01 g/mol

Moles of $CO_2$ = 0.403 g /44.01 g/mol = 0.009157 moles

1 mole of carbon atoms are present in 1 mole of carbon dioxide. So,

Moles of C = 0.009157 moles

Molar mass of C atom = 12.0107 g/mol

Mass of C in molecule = 0.009157 x 12.0107 = 0.11 g

Given that the Valproic acid only contains hydrogen, oxygen and carbon. So,

Mass of O in the sample = Total mass - Mass of C - Mass of H

Mass of the sample = 0.165 g

Mass of O in sample = 0.165 - 0.11 - 0.018588 = 0.036412 g

Molar mass of O = 15.999 g/mol

Moles of O = 0.036412 / 15.999 = 0.002276 moles

Taking the simplest ratio for H, O and C as:

0.01844 : 0.002276 : 0.009157

= 8 : 1 : 4

The empirical formula is = $C_4H_8O$

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 4×12 + 8×1 + 16= 72 g/mol

Molar mass = 144 g/mol

So,

Molecular mass = n × Empirical mass

144 = n × 72

⇒ n = 2

The formula of Valproic acid = $C_8H_{16}O_2$

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