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3 years ago

What is the root mean square of carbon iv oxide at 27°C?

Chemistry

2 answers:

klemol [59]3 years ago

6 0

Answer: not Possible

Explanation:

27 is not. square number

lions [1.4K]3 years ago

5 0

Kinetic Energy (Translational, in all directions) of a gas as per Kinetic theory of gases at abs. Temperature T:

per molecule: KE = 3/2 * k_B T

per mole: KE = 3/2 * R * T

k_B = Boltzmann's constant. R= Universal gas constant.

Kinetic Energy of a gas (as per mechanics)

KE = 1/2 m (Vrms)²

or KE = 1/2 M (Vrms)²

m = mass of a molecule, M = Molar mass of the gas.

Hence, (Vrms)² = 3 k_B T /m = 3 R T /M

R = 8.3 J/°°K/mole

M = 44 gm/mole for CO2 = 0.044 kg/mole

(Vrms)² = 3 * 8.314 J/°K/mole * (273+27)° / (0.044 kg/mole)

Vrms = 412.38 m/sec

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When 125 grams of FeO react with 25.0 grams of Al, how many grams of Fe can be produced? FeO + Al → Fe + Al2O3 25.9 g Fe 38.7 g

Serga [27]

Answer: The mass of iron produced will be 77.6 grams

Explanation:

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

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

For FeO:

Given mass of FeO = 125 g

Molar mass of FeO = 71.8 g/mol

Putting values in equation 1, we get:

$\text{Moles of FeO}=\frac{125g}{71.8g/mol}=1.74mol$

For aluminium:

Given mass of aluminium = 25.0 g

Molar mass of aluminium = 27 g/mol

Putting values in equation 1, we get:

$\text{Moles of aluminium}=\frac{25.0g}{27g/mol}=0.93mol$

The given chemical reaction follows:

$3FeO+2Al\rightarrow 3Fe+Al_2O_3$

By Stoichiometry of the reaction:

2 moles of aluminium metal reacts with 3 mole of FeO

So, 0.93 moles of aluminium metal will react with = $\frac{3}{2}\times 0.93=1.395mol$ of FeO

As, given amount of FeO is more than the required amount. So, it is considered as an excess reagent.

Thus, aluminium metal is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of aluminium metal produces 3 mole of iron metal

So, 0.93 moles of aluminium metal will produce = $\frac{3}{2}\times 0.93=1.395moles$ of iron metal

Now, calculating the mass of iron metal from equation 1, we get:

Molar mass of iron = 55.85 g/mol

Moles of iron = 1.395 moles

Putting values in equation 1, we get:

$1.395mol=\frac{\text{Mass of iron}}{55.85g/mol}\\\\\text{Mass of iron}=(1.395mol\times 55.85g/mol)=77.6g$

Hence, the mass of iron produced will be 77.6 grams

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3 years ago

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