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Mice21 [21]
3 years ago
10

The vapor pressure of ethanol at 25°C is 7.828 kPa . Calculate the vapor pressure in atm .

Chemistry
1 answer:
marshall27 [118]3 years ago
8 0
<span>First you get 7.83 kPa = 7.83 * 10^4 Pa Then convert it to atm by using the relation 1 Pa = 9.8692 * 10^-6 atm, this gives you 7.83 kPa = 7.83 * 10^4 * 9.8692 * 10^-6 atm = 77.275836 * 10^-2 atm = 0.77275836 atm Now round to three significant digits, = 0.773 atm</span>
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What is the mass of the solid NH4Cl formed when 75.5 g of NH3 is mixed with an equal mass of HCl? What is the volume of the gas
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Answer : The volume of the gas remaining is 56.5 liters.

The gas is hydrochloric acid and the formula of the gas is HCl.

The mass of NH_4Cl produced is, 110.7 grams.

Explanation :

The balanced chemical reaction will be:

NH_3+HCl\rightarrow NH_4Cl

First we have to calculate the moles of NH_3 and HCl

\text{Moles of }NH_3=\frac{\text{Mass of }NH_3}{\text{Molar mass of }NH_3}

Molar mass of NH_3 = 17 g/mole

\text{Moles of }NH_3=\frac{75.5g}{17g/mole}=4.44mole

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\text{Moles of }HCl=\frac{\text{Mass of }HCl}{\text{Molar mass of }HCl}

Molar mass of HCl = 36.5 g/mole

\text{Moles of }HCl=\frac{75.5g}{36.5g/mole}=2.07mole

Now we have to calculate the limiting and excess reagent.

From the balanced reaction we conclude that

As, 1 mole of HCl react with 1 mole of NH_3

So, 2.07 mole of HCl react with 2.07 mole of NH_3

From this we conclude that, NH_3 is an excess reagent because the given moles are greater than the required moles and HCl is a limiting reagent and it limits the formation of product.

The remaining moles of HCl gas = 4.44 - 2.07 = 2.37 moles

Now we have to calculate the volume of the gas remaining.

Using ideal gas equation :

PV = nRT

where,

P = Pressure of gas = 752 mmHg = 0.989 atm     (1 atm = 760 mmHg)

V = Volume of gas = ?

n = number of moles of gas = 2.37 moles

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of gas = 14.0^oC=273+14.0=287K

Putting values in above equation, we get:

0.989atm\times V=2.37mole\times (0.0821L.atm/mol.K)\times 287K

V = 56.5 L

Now we have to calculate the moles of NH_4Cl

As, 1 mole of HCl react with 1 mole of NH_4Cl

So, 2.07 mole of HCl react with 2.07 mole of NH_4Cl

Now we have to calculate the mass of NH_4Cl

\text{ Mass of }NH_4Cl=\text{ Moles of }NH_4Cl\times \text{ Molar mass of }NH_4Cl

Molar mass of NH_4Cl = 53.5 g/mole

\text{ Mass of }NH_4Cl=(2.07moles)\times (53.5g/mole)=110.7g

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