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3 years ago

3) In the reaction below, how many grams of carbon dioxide are produced when iron III

Chemistry

1 answer:

Yanka [14]3 years ago

3 0

<h3>Answer:</h3>

132.03 g

<h3>Explanation:</h3>

<u>We are given;</u>

The equation for the reaction as;

Fe₂O₃ + 3CO → 2Fe + 3CO₂

Molar masses of CO and CO₂ as 28.01 g/mol and 44.01 g/mol respectively
Mass of CO as 84 grams

We are required to calculate the mass of CO₂ that will produced.

<h3>Step 1: Calculate the number of moles of CO</h3>

Moles = Mass ÷ Molar mass

Molar mass of CO = 28.01 g/mol

Therefore;

Moles of CO = 84 g ÷ 28.01 g/mol

= 2.9989 moles

= 3.0 moles

<h3>Step 2: Calculate the number of moles of CO₂</h3>

From the reaction, 3 moles of CO reacts to produce 3 moles of CO₂

Therefore; the mole ratio of CO to CO₂ is 1 : 1
Hence; Moles of CO = Moles of CO₂

Moles of CO₂ = 3.0 Moles

But; mass = Moles × molar mass

Thus, mass of CO₂ = 3.0 moles × 44.01 g/mol

= 132.03 g

Hence, the mass of CO₂ produced from the reaction is 132.03 g

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A method used by the U.S. Environmental Protection Agency (EPA) for determining the concentration of ozone in air is to pass the

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Answer: 1. $9.08\times 10^{-6}$ moles

2. 90 mg

Explanation:

$O_3(g)+2NaI(aq)+H_2O(l) \rightarrow O_2(g)+I_2(s)+2NaOH(aq)$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus $4.54 \times 10^{-6}$ moles of ozone is removed by = $\frac{2}{1}\times 4.54 \times 10^{-6}=9.08\times 10^{-6}$ moles of sodium iodide.

Thus $9.08\times 10^{-6}$ moles of sodium iodide are needed to remove $4.54\times 10^{-6}$ moles of $O_3$

2. $\text{Number of moles of ozone}=\frac{0.01331g}{48g/mol}=0.0003moles$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 0.0003 moles of ozone is removed by = $\frac{2}{1}\times 0.0003=0.0006$ moles of sodium iodide.

Mass of sodium iodide= $moles\times {\text {molar mass}}=0.0006\times 150g/mol=0.09g=90mg$ (1g=1000mg)

Thus 90 mg of sodium iodide are needed to remove 13.31 mg of $O_3$ .

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3 years ago

would a cloud or a rock be more dense? PLSSSSSS HELP PLSSS I WILL MARK BRAINLIEST AND ANSWER SOME OF YOUR QUESTIONS

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A rock is definitely more dense. If you were to put a cloud in water it would float/stay above it and a rock would sink to the bottom

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The electrons in the outer shell of an atom are called:<br> outer<br> ionized<br> polar<br> valence

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Valence.

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<span>Heterogeneous:
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3 years ago

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When molten sulfur reacts with chlorine gas, a vile-smelling orange liquid forms that has an empirical formula of SCl. The struc

egoroff_w [7]

Answer:

The structure is shown below.

Explanation:

The formal charge (FC) is the charge that is more close to the actual charge in the real molecules and ions. It can be calculated based on the number of valence electrons (V), the shared electrons (S) and the electrons in the lone pairs (L) by the equation:

FC = V - (L + S/2)

Sulfur is in group 16 of the periodic table, so it has 6 valence electrons, and chlorine is from group 17 of the periodic table, and so it has 7 valence electrons. Chlorine can share only one electron, so it is stable. Sulfur can expand its octet (because it's from the third period) and can have more than 8 electrons when stable.

The possible formulas, from the empiric one, are:

SCl, S₂Cl₂, and S₃Cl₃.

To have FC = 0, chlorine must done only one bond, because S = 2, and L = 6, so:

FC = 7 - (6 + 2/2) = 0

So, it can not be the central atom of a structure. In the SCl, it will hav only a simple bond, so for sulfur, S = 2, and L = 4 (only the lone pairs are counted)

FC = 6 - (4+ 2/2) = +1

For S₂Cl₂, the two sulfurs must be bonded to a simple bond, and each one to one chlorine, thus, for both od them S = 4, and L = 4. so

FC = 6 - (4 + 4/2) = 0

So, it is the correct structure. The lewis structure represents the bonds by lines and the lone pairs of electrons by dots, and it is shown below.

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