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3 years ago

For the following chemical reaction, if 40.0 grams of O2 are completely reacted, how many grams of CO2 will be produced?

Chemistry

1 answer:

e-lub [12.9K]3 years ago

7 0

Hey there!:

Molar mass:

O2 = 31.99 g/mol ; CO2 = 44.01 g/mol

Given the reaction:

C2H4 + 3 O2 --> 2 CO2 + 2 H2O

3 * 31.99 g O2 ------------ 2 * 44.01 g CO2

40.0 g O2 ------------------- mass of CO2 ??

mass of CO2 = 40.0 * 2 * 44.01 / 3 * 31.99

mass of CO2 = 3520.8 / 95.97

mass of CO2 = 36.68 g

Hope that helps!

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In a blast furnace, elemental iron is produced from a mixture of coke (c), iron ore (fe3o4), and other reactants. an important r

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Reaction sequence:

2c(s) + o2(g) -> 2co(g)

fe3o4(s) + 4co(g) -> 3fe(l) + 4co2(g)

According to first equation, 2 moles of carbon produce 2 moles of carbon monoxide. So 1 mole of carbon will produce 1 mol of carbon monoxide (the same number).

According to the second equation, 4 moles of carbon monoxide produce 3 moles of iron. We should make the cross multiplication with those numbers:

4 moles CO/3 moles iron = 1 mol CO/x

x = 1 mol CO*3 moles iron/4 moles CO = 0.75 moles of iron

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3 years ago

According to the periodic table, the neutral atom of what element has 4

liberstina [14]

Answer:

Beryllium

Explanation:

Given parameters:

Number of protons = 4

Number of neutrons = 5

Number of electrons = 4

Unknown:

Element = ?

Solution:

To determine an element, the atomic number is the most important identifier.

The atomic number is the number of protons in an atom.

Since the atom here has a proton number of 4, checking on the periodic table, it is Beryllium, Be

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3 years ago

Is Fe2+ + 2e- = Fe an oxidation or reduction

Law Incorporation [45]

Answer:

oxidation

Explanation:

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3 years ago

A hypothetical element has an atomic weight of 48.68 amu. It consists of three isotopes having masses of 47.00 amu, 48.00 amu, a

Morgarella [4.7K]

Answer : The percent abundance of the heaviest isotope is, 78 %

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$

As we are given that,

Average atomic mass = 48.68 amu

Mass of heaviest-weight isotope = 49.00 amu

Let the percentage abundance of heaviest-weight isotope = x %

Fractional abundance of heaviest-weight isotope = $\frac{x}{100}$

Mass of lightest-weight isotope = 47.00 amu

Percentage abundance of lightest-weight isotope = 10 %

Fractional abundance of lightest-weight isotope = $\frac{10}{100}$

Mass of middle-weight isotope = 48.00 amu

Percentage abundance of middle-weight isotope = [100 - (x + 10)] % = (90 - x) %

Fractional abundance of middle-weight isotope = $\frac{(90-x)}{100}$

Now put all the given values in above formula, we get:

$48.68=[(47.0\times \frac{10}{100})+(48.0\times \frac{(90-x)}{100})+(49.0\times \frac{x}{100})]$

$x=78\%$

Therefore, the percent abundance of the heaviest isotope is, 78 %

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