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STALIN [3.7K]
2 years ago
5

A 0.500-g sample containing Na2CO3 plus inert matter is analyzed by adding 50.0 mL of 0.100 M HCl, a slight excess, boiling to r

emove CO2, and then back-titrating the excess acid with 0.100 M NaOH. If 5.6 mL NaOH is required for the back-titration, what is the percent Na2CO3 in the sample?
Chemistry
1 answer:
Yakvenalex [24]2 years ago
5 0

The percentage by mass of Na2CO3 in the sample is 48%.

The equation of the reaction of Na2CO3 with HCl;

Na2CO3(aq) + 2HCl(aq) ------> 2NaCl(aq) + H2O(l) + CO2(g)

Since the HCl is in excess, the excess is back titrated with NaOH as follows;

NaOH (aq) + HCl(aq) ---->NaCl(aq) + H2O(l)

Number of moles of HCl added =  0.100 M × 50/1000 L = 0.005 moles

Number of moles of NaOH added = 5.6/1000 ×  0.100 M = 0.00056 moles

Since the reaction of NaOH and NaOH is 1:1, 0.00056 moles of HCl reacted with excess HCl.

Amount of HCl that reacted with Na2CO3 =  0.005 moles -  0.00056 moles = 0.0044 moles

Now;

1 mole of Na2CO3 reacts with 2 moles of HCl

x moles of Na2CO3 reacts with 0.0044 moles of HCl

x = 1 mole × 0.0044 moles / 2 moles

x = 0.0022 moles

Mass of Na2CO3 reacted = 0.0022 moles × 106 g/mol = 0.24 g

Percentage of Na2CO3  in the sample = 0.24 g/ 0.500-g × 100/1 = 48%

Learn more about back titration: brainly.com/question/25485091

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The molarity of a 4.200 L solution is 1.230 M Na2CO3. What is the mass of Na2CO3
Lerok [7]

Answer:

547.5g

Explanation:

To get the mass, you need moles.

moles = (molarity)(Liters)

moles = (1.230M)(4.200L) = 5.166 moles Na2CO3

Now, just use stoichiometry

molar mass of Na2CO3 = 2(mass of Na) + (mass of C) + 3(mass of O)

= 2(22.9) + 12.01 + 3(16) = 105.99g/mol

5.166moles(105.99g/mol)

= 547.544

But, the measurements given had 4 significant figures, so in chemistry we write:

547.5g

7 0
3 years ago
Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/m
koban [17]

Here is the full question:

Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.2% carbon dioxide, determine the subsequent amount in the room at any time.

What is the concentration at 10 minutes? (Round your answer to three decimal places.

Answer:

0.046 %

Explanation:

The rate-in;

R_{in} = \frac{0.04}{100}*2000

R_{in} = 0.8

The rate-out

R_{out} = \frac{A}{6000}*2000

R_{out} = \frac{A}{3}

We can say that:

\frac{dA}{dt}= 0.8-\frac{A}{3}

where;

A(0)= 0.2% × 6000

A(0)= 0.002 × 6000

A(0)= 12

\frac{dA}{dt} +\frac{A}{3} =0.8

Integration of the above linear equation =

e^{\int\limits \frac {1}{3}dt } = e^{\frac{1}{3}t

so we have:

e^{\frac{1}{3}t}\frac{dA}{dt}} +\frac{1}{3}e^{\frac{1}{3}t}A = 0.8e^{\frac{1}{3}t

\frac{d}{dt}[e^{\frac{1}{3}t}A] = 0.8e^{\frac{1}{3}t

Ae^{\frac{1}{3}t} =2.4e\frac{1}{3}t +C

∴ A(t) = 2.4 +Ce^{-\frac{1}{3}t

Since A(0) = 12

Then;

12 =2.4 + Ce^{-\frac{1}{3}}(0)

C= 12-2.4

C =9.6

Hence;

A(t) = 2.4 +9.6e^{-\frac{t}{3}}

A(0) = 2.4 +9.6e^{-\frac{10}{3}}

A(t) = 2.74

∴ the concentration at 10 minutes is ;

=  \frac{2.74}{6000}*100%

= 0.0456667 %

= 0.046% to three decimal places

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It is the mass of one mole of the substance.
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ycow [4]

Answer:

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2 years ago
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notka56 [123]

Answer:

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Every colloid consists of two parts :colloidal particles and the dispersing medium.

3 0
3 years ago
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