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Ratling [72]
3 years ago
10

Help me with 4,5 plz

Chemistry
1 answer:
Helen [10]3 years ago
7 0
The answer to 4 is A.                                              
The answer to 5 is C.
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Dissolving 5.28 g of an impure sample of calcium carbonate in hydrochloric acid produced 1.14 L of carbon dioxide at 20.0 °C and
swat32

Answer:

\%\ mass\ of\ CaCO_3=93.37\ \%

Explanation:

Given that:

Pressure = 791 mmHg

Temperature = 20.0°C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (20 + 273.15) K = 293.15 K  

T = 293.15 K  

Volume = 100 L

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 62.3637 L.mmHg/K.mol  

Applying the equation as:

791 mmHg × 1.14 L = n × 62.3637 L.mmHg/K.mol  × 293.15 K  

⇒n of CO_2 produced =  0.0493 moles

According to the reaction:-

CaCO_3 + 2 HCl\rightarrow CaCl_2 + H_2O + CO_2

1 mole of carbon dioxide is produced 1 mole of calcium carbonate reacts

0.0493 mole of carbon dioxide is produced 0.0493 mole of calcium carbonate reacts

Moles of calcium carbonate reacted = 0.0493 moles

Molar mass of CaCO_3 = 100.0869 g/mol

The formula for the calculation of moles is shown below:

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

Thus,

0.0493\ mol= \frac{Mass}{100.0869\ g/mol}

Mass_{CaCO_3}=4.93\ g

Impure sample mass = 5.28 g

Percent mass is percentage by the mass of the compound present in the sample.

\%\ mass\ of\ CaCO_3=\frac{Mass_{CaCO_3}}{Total\ mass}\times 100

\%\ mass\ of\ CaCO_3=\frac{4.93}{5.28}\times 100

\%\ mass\ of\ CaCO_3=93.37\ \%

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