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nalin [4]
2 years ago
10

Hydrogen chloride gas can be prepared by the following reaction: 2NaCl(s) + H2SO4(aq) → 2HCl(g) + Na2SO4(s) How many grams of HC

l can be prepared from 2.00 mol H2SO4 and 2.56 mol NaCl?

Chemistry
1 answer:
SVETLANKA909090 [29]2 years ago
7 0

Answer:

The correct answer is is option B

b. 93.3 g

Explanation:

SEE COMPLETE QUESTION BELOW

Hydrogen chloride gas can be prepared by the following reaction: 2NaCl(s) + H2SO4(aq) → 2HCl(g) + Na2SO4(s)

How many grams of HCl can be prepared from 2.00 mol H2SO4 and 2.56 mol NaCl?

a. 7.30 g

b. 93.3 g

c. 146 g

d. 150 g

e. 196 g

CHECK THE ATTACHMENT FOR STEP BY STEP EXPLANATION

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Answer:

The molar mass of the compound is:- 168.82 g/mol

The molar mass of the gas is:- 16.38 g/mol

Explanation:

(a)

Using ideal gas equation as:

PV=nRT

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P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Also,  

Moles = mass (m) / Molar mass (M)

Density (d)  = Mass (m) / Volume (V)

So, the ideal gas equation can be written as:

PM=dRt

Given that:-

Pressure = 20 kPa = 20000 Pa

The expression for the conversion of pressure in Pascal to pressure in atm is shown below:

P (Pa) = \frac {1}{101325} P (atm)

20000 Pa = \frac {20000}{101325} atm

Pressure = 0.1974 atm

Temperature = 330 K

d = 1.23 kg/m³ = 1.23 g/L

Molar mass = ?

Applying the equation as:

0.1974 atm × M = 1.23 g/L × 0.0821 L.atm/K.mol × 330 K

⇒M = 168.82 g/mol

<u>The molar mass of the compound is:- 168.82 g/mol</u>

(b)

Given that:

Pressure = 152 Torr

Temperature = 298 K

Volume = 250 cm³ = 0.25 L

Using ideal gas equation as:

PV=nRT

R = 62.3637\text{torr}mol^{-1}K^{-1}

Applying the equation as:

152 Torr × 0.25 L = n × 62.3637 L.torr/K.mol × 298 K

⇒n = 0.002045 moles

Given that :  

Mass of the gas = 33.5 mg = 0.0335 g

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The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

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0.002045\ moles&#10;= \frac{0.0335\ g}{Molar\ mass}

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Answer:

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Explanation:

Hello,

In this case, given that the mass of the product is 0.534 g, we can infer that the percent composition of tin is:

\%Sn=\frac{0.500g}{0.534g}*100\%\\ \\\%Sn=93.6\%

Therefore, the percent composition of oxygen is 6.4% for a 100% in total. Thus, with such percents we compute the moles of each element in the oxide:

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In such a way, for finding the smallest whole number we divide the moles of both tin and oxygen by the moles of oxygen as the smallest moles:

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Sn_2O

Best regards.

8 0
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