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2 years ago

Calculate how many grams of iron can be made from 16.5 grams of iron(III) oxide if hydrogen gas is in excess?

1 answer:

7 0

Taking into account the reaction stoichiometry, 11.54 grams of Fe are formed from 16.5 grams of iron(III) oxide if hydrogen gas is in excess.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

Fe₂O₃ + 3 H₂ → 2 Fe + 3 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Fe₂O₃: 1 mole
H₂: 3 moles
Fe: 2 moles
H₂O: 3 moles

The molar mass of the compounds is:

Fe₂O₃: 159.7 g/mole
H₂: 2 g/mole
Fe: 55.85 g/mole
H₂O: 18 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

Fe₂O₃: 1 mole ×159.7 g/mole= 159.7 grams
H₂: 3 moles ×2 g/mole= 6 grams
Fe: 2 moles ×55.85 g/mole= 111.7 grams
H₂O: 3 moles ×18 g/mole= 54 grams

<h3>Mass of each product formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 159.7 grams of Fe₂O₃ form 111.7 grams of Fe, 16.5 grams of Fe₂O₃ form how much mass of Fe?

$mass of Fe=\frac{16.5 grams of Fe_{2} O_{3} x111.7 grams of Fe}{159.7 grams of Fe_{2} O_{3}}$

<u><em>mass of Fe= 11.54 grams</em></u>

Then, 11.54 grams of Fe are formed from 16.5 grams of iron(III) oxide if hydrogen gas is in excess.

Learn more about the reaction stoichiometry:

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A 0.187 M weak acid solution has a pH of 3.99. Find Ka for the acid. Express your answer using two significant figures.

Anna35 [415]

Answer:

5.56 × 10⁻⁸

Explanation:

Step 1: Given data

Concentration of the weak acid (Ca): 0.187 M

pH of the solution: 3.99

Step 2: Calculate the concentration of H⁺

We will use the following expression.

pH = -log [H⁺]

[H⁺] = antilog -pH = antilog -3.99 = 1.02 × 10⁻⁴ M

Step 3: Calculate the acid dissociation constant (Ka)

We will use the following expression.

$Ka = \frac{[H^{+}]^{2} }{Ca} = \frac{(1.02 \times 10^{-4})^{2} }{0.187} = 5.56 \times 10^{-8}$

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2 years ago

Answer these questions based on 234. 04360 as the atomic mass of thorium-234. The masses for the subatomic particles are given.

nikdorinn [45]

The mass defect for the isotope thorium-234 if given mass is 234.04360 amu is 1.85864 amu.

<h3>How do we calculate atomic mass?</h3>

Atomic mass (A) of any atom will be calculated as:

A = mass of protons + mass of neutrons

In the Thorium-234:

Number of protons = 90

Number of neutrons = 144

Mass of one proton = 1.00728 amu

Mass of one neutron = 1.00866 amu

Mass of thorium-234 = 90(1.00728) + 144(1.00866)

Mass of thorium-234 = 90.6552 + 145.24704 = 235.90224 amu

Given mass of thorium-234 = 234.04360 amu

Mass defect = 235.90224 - 234.04360 = 1.85864 amu

Hence required value is 1.85864 amu.

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1 year ago

How much energy is required to melt 10. 0 g of ice at 0. 0°C, warm it to 100. 0°c and completely vaporize the sample?

natta225 [31]

The 7160 cal energy is required to melt 10. 0 g of ice at 0. 0°C, warm it to 100. 0°C and completely vaporize the sample.

Calculation,

Given data,

Mass of the ice = 10 g

Temperature of ice = 0. 0°C

The ice at 0. 0°C is to be converted into water at 0. 0°C

Heat required at this stage = mas of the ice ×latent heat of fusion of ice

Heat required at this stage = 10 g×80 = 800 cal

The temperature of the water is to be increased from 0. 0°C to 100. 0°C

Heat required for this = mass of the ice×rise in temperature×specific heat of water

Heat required for this = 10 g×100× 1 = 1000 cal

This water at 100. 0°C is to be converted into vapor.

Heat required for this = Mass of water× latent heat

Heat required for this = 10g ×536 =5360 cal

Total energy or heat required = sum of all heat = 800 +1000+ 5360 = 7160 cal

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1 year ago

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kondaur [170]

Answer:

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Explanation:

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2 years ago

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