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The Solvay process for the manufacture of sodium carbonate begins by passing ammonia and carbon dioxide through a solution of sodium chloride to make sodium bicarbonate and ammonium chloride. The equation for this reaction is H2O + NaCl + NH3 + CO2 �¨ NH4Cl + NaHCO3. In the next step, sodium bicarbonate is heated to give sodium carbonate and two gases, carbon dioxide and steam 2NaHCO3 �¨ Na2CO3 + CO2 + H2O.

What is the theoretical yield of sodium carbonate, expressed in grams, if 157 g of NaCl were used in the first reaction?---------

If 93.8 g of Na2CO3 were obtained from the reaction described in part a, what was the percentage yield?---------

Explanation please

User Icon for DrBob222 DrBob222 answered
14 years ago

You have the balanced equations.

2. Convert 157 g NaCl to moles. moles = grams/molar mass.
3. Using the coefficients in the balanced equation, convert moles NaCl to moles Na2CO3.
4. Now convert moles Na2CO3 to grams. g = moles x molar mass. This the theoretical yield.
5. %yield = (actual yield/theoretical yield)*100 = ??

The problem gives 93.8 g Na2CO3 as the actual yield.

User Icon for Step-by-Step Bot Step-by-Step Bot answered
1 year ago

To calculate the theoretical yield of sodium carbonate, we need to determine the stoichiometry of the reaction. From the balanced equation:

H2O + NaCl + NH3 + CO2 → NH4Cl + NaHCO3

We can see that the molar ratio of NaCl to NaHCO3 is 1:1.

First, we convert the given mass of NaCl to moles using the molar mass of NaCl:
157 g NaCl × (1 mol NaCl / 58.44 g NaCl) = 2.69 mol NaCl

Since the ratio of NaCl to NaHCO3 is 1:1, the number of moles of NaHCO3 produced is also 2.69 mol.

Next, we look at the second reaction:

2NaHCO3 → Na2CO3 + CO2 + H2O

From the balanced equation, we can determine the stoichiometry between NaHCO3 and Na2CO3: 2:1.

Therefore, the number of moles of Na2CO3 produced is 2.69 mol × (1 mol Na2CO3 / 2 mol NaHCO3) = 1.345 mol Na2CO3.

Finally, we calculate the mass of Na2CO3 using the molar mass of Na2CO3:
1.345 mol Na2CO3 × (105.99 g Na2CO3 / 1 mol Na2CO3) = 143 g Na2CO3

So, the theoretical yield of sodium carbonate is 143 grams.

To calculate the percentage yield, we use the formula:

Percentage Yield = (Actual Yield / Theoretical Yield) × 100

Given that the actual yield of Na2CO3 is 93.8 g, we substitute the values into the formula:

Percentage Yield = (93.8 g Na2CO3 / 143 g Na2CO3) × 100 = 65.7%

Therefore, the percentage yield of sodium carbonate is 65.7%.

User Icon for Explain Bot Explain Bot answered
1 year ago

To determine the theoretical yield of sodium carbonate in grams, you need to use stoichiometry and the balanced equation of the reaction.

First, we need to calculate the molar mass of NaCl. The molar mass of Na is 22.99 g/mol, and the molar mass of Cl is 35.45 g/mol. So the molar mass of NaCl is:

(22.99 g/mol Na) + (35.45 g/mol Cl) = 58.44 g/mol NaCl

Next, we need to convert the mass of NaCl used to moles. We can do this by dividing the given mass by the molar mass:

157 g NaCl / 58.44 g/mol NaCl = 2.69 mol NaCl

Now, let's examine the balanced equation to determine the stoichiometry between NaCl and Na2CO3. From the equation, we see that it takes 1 mole of NaCl to produce 1 mole of Na2CO3. Therefore, the moles of Na2CO3 produced are also 2.69 mol.

Finally, we can calculate the mass of Na2CO3 using the molar mass of Na2CO3. The molar mass of Na2CO3 is:

(22.99 g/mol Na * 2) + (12.01 g/mol C) + (16.00 g/mol O * 3) = 105.99 g/mol Na2CO3

So, the theoretical yield of Na2CO3 is:

2.69 mol Na2CO3 * 105.99 g/mol Na2CO3 = 285 g Na2CO3

Therefore, the theoretical yield of sodium carbonate is 285 grams.

To calculate the percentage yield, we need to use the actual yield and divide it by the theoretical yield, then multiply by 100.

Given: Actual yield = 93.8 g Na2CO3

Percentage yield = (Actual yield / Theoretical yield) * 100
= (93.8 g / 285 g) * 100
= 32.91%

Therefore, the percentage yield of sodium carbonate is approximately 32.91%.