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belka [17]
3 years ago
5

B) How many kilograms of carbon dioxide are formed when 24.42 g of iron is produced?

Chemistry
1 answer:
charle [14.2K]3 years ago
5 0

Answer:

0.0289 kg of CO2 will be formed

Explanation:

Step 1: Data given

Mass of iron produced = 24.42 grams

Atomic mass iron = 55.845 g/mol

Molar mass CO2 = 44.01 g/mol

Step 2: The balanced equation

Fe2O3 + 3CO → 2Fe + 3CO2

Step 3: Calculate moles iron

Moles iron = mass iron / molar mass iron

Moles iron = 24.42 grams / 55.845 g/mol

Moles iron = 0.437 moles

Step 4: Calculate moles CO2

For 1 mol Fe2O3 we need 3 moles CO to produce 2 moles Fe and 3 moles CO2

For 0.437 moles Fe we'll have 3/2 * 0.437 = 0.6555 moles CO2

Step 5: Calculate mass CO2

Mass CO2 = moles CO2 * molar mass CO2

Mass CO2 = 0.6555 moles * 44.01 g/mol

Mass CO2 = 28.85 grams = 0.0289 kg

0.0289 kg of CO2 will be formed

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It would take 147 hours for 320 g of the sample to decay to 2.5 grams from the information provided.

Radioactivity refers to the decay of a nucleus leading to the spontaneous emission of radiation. The half life of a radioactive nucleus refers to the time required for the nucleus to decay to half of its initial amount.

Looking at the table, we can see that the initial mass of radioactive material present is 186 grams, within 21 hours, the radioactive substance decayed to half of its initial mass (93 g). Hence, the half life is 21 hours.

Using the formula;

k = 0.693/t1/2

k = 0.693/21 hours = 0.033 hr-1

Using;

N=Noe^-kt

N = mass of radioactive sample at time t

No = mass of radioactive sample initially present

k = decay constant

t = time taken

Substituting values;

2.5/320= e^- 0.033 t

0.0078 = e^- 0.033 t

ln (0.0078) = 0.033 t

t = ln (0.0078)/-0.033

t = 147 hours

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