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Drupady [299]
3 years ago
5

Consider the reaction. 4 k(s) + o2(g) s 2 k2o(s) the molar mass of k is 39.09 g>mol and that of o 32.00 g>mol. Without doi

ng any calculations, pick the condi- tions under which potassium is the limiting reactant and explain your reasoning.
a. 170 g k, 31 g o2
b. 16 g k, 2.5 g o2
c. 165 kg k, 28 kg o2
d. 1.5 g k, 0.38 g o2
Chemistry
2 answers:
Grace [21]3 years ago
7 0

Answer:

d. 1.5 g k, 0.38 g o2

Explanation:

Scilla [17]3 years ago
3 0

Answer : The correct option is (d) 1.5 g K, 0.38 g O₂

Explanation :

The molar masses of potassium and oxygen are very close to each other. Therefore we can assume them to be equal. If we assume that, then according to reaction stoichiometry, 4 moles of K are needed to react with 1 mol of O₂. Since the molar masses are assumed to be equal , we can say that the mass of potassium needed to react with that of oxygen should be 4 times the mass of oxygen.

From the given options, the only option that has less amount of K is option d.

Here, 0.38 g of oxygen needs 0.38 x 4 = 1.52 g of K. But the given mass of potassium is 1.5 g which is less. This indicates that potassium is the limiting reactant as we do not have enough potassium to completely react with all of the oxygen.

Therefore option d is the correct option

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Number of electrons = (6.7 x 10^22 atoms) * (47 electrons / atom)

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Answer : The new density and new volume of carbon dioxide gas is 0.2281 g/L and 7.2m^3 respectively.

Explanation :

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V_1 = initial volume of gas = 10m^3

V_2 = final volume of gas = ?

T_1 = initial temperature of gas = 50^oC=273+50=323K

T_2 = final temperature of gas = 75^oC=273+75=348K

Now put all the given values in the above equation, we get:

\frac{10kPa\times 10m^3}{323K}=\frac{15kPa\times V_2}{348K}

V_2=7.2m^3

The new volume of carbon dioxide gas is 7.2m^3

Now we have to calculate the new density of carbon dioxide gas.

PV=nRT\\\\PV=\frac{m}{M}RT\\\\P=\frac{m}{V}\frac{RT}{M}\\\\P=\rho \frac{RT}{M}\\\\\rho=\frac{PM}{RT}

Formula for new density will be:

\rho_2=\frac{P_2M}{RT_2}

where,

P_2 = new pressure of gas = 15 kPa

T_2 = new temperature of gas = 75^oC=273+75=348K

M = molar mass of carbon dioxide gas = 44 g/mole

R = gas constant = 8.314 L.kPa/mol.K

\rho = new density

Now put all the given values in the above equation, we get:

\rho_2=\frac{(15kPa)\times (44g/mole)}{(8.314L.kPa/mol.K)\times (348K)}

\rho_2=0.2281g/L

The new density of carbon dioxide gas is 0.2281 g/L

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