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3 years ago

A tank contains 7.0 moles of a mixture of nitrogen gas and oxygen gas. The total

Chemistry

1 answer:

Pavel [41]3 years ago

8 0

Answer:

5.2 mol

Explanation:

Step 1: Given data

Total number of moles (n): 7.0 mol

Total pressure (P): 530 kPa

Partial pressure of oxygen (pO₂): 140 kPa

Step 2: Calculate the mole fraction of oxygen

We will use the following expression.

pO₂ = P × X(O₂)

X(O₂) = pO₂ / P

X(O₂) = 140 kPa / 530 kPa

X(O₂) = 0.264

Step 3: Calculate the moles of oxygen

We will use the definition of mole fraction of oxygen.

X(O₂) = n(O₂) / n

n(O₂) = X(O₂) × n

n(O₂) = 0.264 × 7.0 mol

n(O₂) = 1.8 mol

Step 4: Calculate the moles of nitrogen

The total number of moles is equal to the sum of moles of the individual gases.

n(O₂) + n(N₂) = n

n(N₂) = n - n(O₂)

n(N₂) = 7.0 mol - 1.8 mol

n(N₂) = 5.2 mol

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2 years ago

onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency

mestny [16]

Answer : The excess reactant is, $H_2O$

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of $CaCN_2$ = 105.0 g

Mass of $H_2O$ = 78.0 g

Molar mass of $CaCN_2$ = 80.11 g/mole

Molar mass of $H_2O$ = 18 g/mole

Molar mass of $CaCO_3$ = 100.09 g/mole

First we have to calculate the moles of $CaCN_2$ and $H_2O$ .

$\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles$

$\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3$

From the balanced reaction we conclude that

As, 1 mole of $CaCN_2$ react with 3 mole of $H_2O$

So, 1.31 moles of $CaCN_2$ react with $1.31\times 3=3.93$ moles of $H_2O$

From this we conclude that, $H_2O$ is an excess reagent because the given moles are greater than the required moles and $CaCN_2$ is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

$\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)$

$\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g$

Thus, the leftover amount of excess reagent is, 7.2 grams.

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4 years ago

The specific heat of zinc is 0.39 J/g*°C. How much energy needed to change the temperature of 34g of zinc from 22°C to 57°C. Is

Svetlanka [38]

Answer:

464.1 J absorbed.

Explanation:

Given data:

Specific heat of zinc = 0.39 J/g°C

Mass of zinc = 34 g

Temperature changes = 22°C to 57°C

Energy absorbed or released = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 57°C - 22°C

ΔT = 35°C

Q = m.c. ΔT

Q = 34 g. 0.39 J/g°C. 35°C

Q = 464.1 J

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3 years ago