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nekit [7.7K]
4 years ago
5

Baking soda NaHCO3, is made from soda ash, a common name for sodium carbonate. The soda ash is obtained in two ways. It can be m

anufactured in a process in which carbon dioxide, ammonia, sodium chloride, and water are the starting materials. Alternatively, it is mined as a mineral called trona (top photo). Whether the soda ash is mined or manufactured, it is dissolved in water and carbon dioxide is bubbled through the solution. Sodium bicarbonate precipitates from the solution. As a chemical analyst you are presented with two samples of sodium bicarbonateâone from the manufacturing process and the other derived from trona. You are asked to determine which is purer and are told that the impurity is sodium carbonate. You decide to treat the samples with just sufficient hydrochloric acid to convert all the sodium carbonate and bicarbonate to sodium chloride, carbon dioxide, and water. You then precipitate silver chloride in the reaction of sodium chloride with silver nitrate. A 6.93 g sample of baking soda derived from trona gave 11.89 g of silver chloride. A6.78 g sample from manufactured sodium carbonate gave 11.77 g of silver chloride.
Required:
Which sample is purer, that is, which has the greater mass percent NaHCO3?
Chemistry
1 answer:
Vaselesa [24]4 years ago
8 0

Answer:

Explanation:

From the given information:

mass of silver chloride AgCl = 11.89 g

molar mass of AgCl = 143.37 g/mol

We know that:

number of moles = mass/molar mass

∴

number of moles of AgCl = 11.89 g/ 143.37 g/mol

number of moles of AgCl = 0.0829 mol

The chemical equation for the mineral called trona is:

\mathsf{Na_2CO_3.NaHCO_3.2H_2O}

when being reacted with hydrochloric acid, we have:

\mathsf{Na_2CO_3.NaHCO_3.2H_2O + 3HCl \to 3NaCl + 2CO_2 +4H_2O}

One mole of NaCl formed from one mole of trona sample = 0.0829 moles of AgCl

i.e. 0.0829 moles of NaCl can be formed from AgCl

mass of trona sample = number of moles × molar mass

mass of trona sample = 0.0829 × 226

mass of trona sample = 18.735 g

The mass in the percentage of NaHCO₃ = mass of NaHCO₃/ mass of trona

The mass in the percentage of NaHCO₃ = 6.93/18.735

The mass in the percentage of NaHCO₃ = 0.36989

The mass in the percentage of NaHCO₃ = 36.99%

Nonetheless, a 6.78 g samples manufactured from sodium carbonate in pure 100%

∴

6.78 g sample manufactured from Na₂CO₃ is purer.

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A solution labeled "0.105 M NaOH" would contain ______________ moles of NaOH in each liter of solution.
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When heated to 150 ºC, CuSO4.5 H2O loses its water of hydration as gaseous H2O. A 2.50 g sample of the compound is placed in a s
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Answer:

Water pressure 0.5 atm

Total Pressure= 2.27  atm

Explanation:

To answer this problem, one has to realize that there are two processes that increase the temperature of the sealed vessel.

First, the dry air in the sealed vessel will be heated which will cause its pressure to increase and it can be determined by the equation:

P₁ x T₂   = P₂ x T₁  ∴  P₂ =  P₁ x T₂ / T₁

For the second process, we have an amount of n moles of water which will be released when the copper sulfate is heated. In this case, to determine the value of the the water gas we will use the gas law:

   PV = nRT  ∴ P =  nRT/V

n will we calculated from the quantity of sample.

2.50 g  CuSo₄ 5H₂O x  1 mol/ 249.69 g = 0.01 mol CuSo₄ 5H₂O

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= 0.05 mo H₂O

pressure of dry air at the final temperature,

P₂ = 1 atm x 500 K/ 300 K = 1.67 atm

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A bomb calorimeter has a heat capacity of 675 J/°C and contains 925 g of water. If the combustion of 0.500 mole of a hydrocarbon
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<u>Answer:</u> The enthalpy of the reaction is 269.4 kJ/mol

<u>Explanation:</u>

To calculate the heat absorbed by the calorimeter, we use the equation:

q_1=c\Delta T

where,

q = heat absorbed

c = heat capacity of calorimeter = 675 J/°C

\Delta T = change in temperature = T_2-T_1=(53.88-24.26)^oC=29.62^oC

Putting values in above equation, we get:

q_1=675J/^oC\times 29.62^oC=19993.5J

To calculate the heat absorbed by water, we use the equation:

q_2=mc\Delta T

where,

q = heat absorbed

m = mass of water = 925 g

c = heat capacity of water = 4.186 J/g°C

\Delta T = change in temperature = T_2-T_1=(53.88-24.26)^oC=29.62^oC

Putting values in above equation, we get:

q_2=925g\times 4.186J/g^oC\times 29.62^oC=114690.12J

Total heat absorbed = q_1+q_2

Total heat absorbed = [19993.5+114690.12]J=134683.62J=134.7kJ

To calculate the enthalpy change of the reaction, we use the equation:

\Delta H_{rxn}=\frac{q}{n}

where,

q = amount of heat absorbed = 134.7 kJ

n = number of moles of hydrocarbon = 0.500 moles

\Delta H_{rxn} = enthalpy change of the reaction

Putting values in above equation, we get:

\Delta H_{rxn}=\frac{134.7kJ}{0.500mol}=269.4kJ/mol

Hence, the enthalpy of the reaction is 269.4 kJ/mol

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Learn more about the Nitrate reduction test with the help of the given link:

brainly.com/question/11181586

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