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4 years ago

A 32 L samples of xenon gas at 10°C is expanded to 35 L. Calculate the final temperature.

Chemistry

1 answer:

morpeh [17]4 years ago

3 0

Answer:

Final Temperature = 36.54 ⁰C

Explanation:

Lets suppose the gas is acting ideally, then according to Charle's Law, "The volume of a fixed mass of gas at constant pressure is directly proportional to the absolute temperature". Mathematically for initial and final states the relation is as follow,

V₁ / T₁ = V₂ / T₂

Data Given;

V₁ = 32 L

T₁ = 10 °C = 283.15 K ∴ K = °C + 273.15

V₂ = 35 L

T₂ = ??

Solving equation for T₂,

T₂ = V₂ × T₁ / V₁

Putting values,

T₂ = (35 L × 283.15 K) ÷ 32 L

T₂ = 309.69 K ∴ ( 36.54 °C )

Result:

As the volume is increased from 32 L to 35 L, therefore, the temperature must have increased from 10 °C to 36.54 °C.

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Answer:

For A: The mass of aluminium required will be 183 g

For B: The mass of alumina produced will be $6.63\times 10^6g$

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For a:

Given mass of $NH_4ClO_4$ = 1.325 kg = 1325 g (Conversion factor: 1 kg = 1000 g)

Molar mass of $NH_4ClO_4$ = 117.50 g/mol

Putting values in equation 1, we get:

$\text{Moles of }NH_4ClO_4=\frac{1325g}{117.50g/mol}=11.28mol$

For the given chemical reaction:

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By stoichiometry of the reaction:

6 moles of $NH_4ClO_4$ reacts with 10 moles of aluminium

So, 11.28 moles of $NH_4ClO_4$ will react with = $\frac{10}{6}\times 11.28=6.77mol$ of aluminium

Now, calculating the mass of aluminium by using equation 1, we get:

Molar mass of aluminium = 27.00 g/mol

Moles of aluminium = 6.77 moles

Putting values in equation 1, we get:

$6.77mol=\frac{\text{Mass of aluminium}}{27.00g/mol}\\\\\text{Mass of aluminium}=(6.77mol\times 27.00g/mol)=183g$

Hence, the mass of aluminium required will be 183 g

For b:

Given mass of aluminium = $3.500\times 10^3=3.5\times 10^6g$ (Conversion factor: 1 kg = 1000 g)

Molar mass of aluminium = 27.00 g/mol

Putting values in equation 1, we get:

$\text{Moles of aluminium}=\frac{3.5\times 10^6g}{27.00g/mol}=1.3\times 10^5mol$

For the given chemical reaction:

$6NH_4ClO_4(s)+10Al(s)\rightarrow 5Al_2O_3(s)+3N_2(g)+6HCl(g)+9H_2O(g)$

By stoichiometry of the reaction:

10 moles of aluminium produces 5 moles of alumina

So, $1.3\times 10^5mol$ of aluminium will react with = $\frac{5}{10}\times 1.3\times 10^5=6.5\times 10^4mol$ of alumina

Now, calculating the mass of alumina by using equation 1, we get:

Molar mass of alumina = 101.96 g/mol

Moles of alumina = $6.5\times 10^4mol$

Putting values in equation 1, we get:

$6.5\times 10^4mol=\frac{\text{Mass of alumina}}{101.96g/mol}\\\\\text{Mass of alumina}=(6.5\times 10^4mol\times 101.96g/mol)=6.63\times 10^6g$

Hence, the mass of alumina produced will be $6.63\times 10^6g$

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