Chemistry, 25.02.2020 05:27 theoriginalstal4234

The mass of sodium chloride in (g) is 14.19 The volume of ammonia solution in (mL) is 36.15 Calculate the following: What is the theoretical yield of sodium bicarbonate in grams?

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Answer from: hma9153

This is an incomplete question, here is a complete question.

A weighed amount of sodium chloride is completely dissolved in a measured volume of 4.00 M ammonia solution at ice temperature, and carbon dioxide is bubbled in. Assume that sodium bicarbonate is formed util the limiting reagent is entirely used up. the solubility of sodium bicarbonate in water at ice temperature is 0.75 mol per liter. Also assume that all the sodium bicarbonate precipitated is collected and converted quantitatively to sodium carbonate.

Data to be used for calculating the results

-The mass of sodium chloride in (g) is 14.19

-The volume of ammonia solution in (mL) is 36.15

Calculate the following: What is the theoretical yield of sodium bicarbonate in grams?

Answer : The theoretical yield of sodium bicarbonate in grams is, 20.4 grams.

Explanation :

First we have to calculate the moles of NaCl and NH_3 .

$\text{ Moles of }NaCl=\frac{\text{ Mass of }NaCl}{\text{ Molar mass of }NaCl}=\frac{14.19g}{58.5g/mole}=0.243moles$

$\text{ Moles of }NH_3=\text{ Concentration of }NH_3\times \text{ Volume of solution}=4.00M\times 0.3615L=1.446moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction will be:

$NH_3+NaCl+CO_2+H_2O\rightarrow NaHCO_3+NH_4Cl$

From the balanced reaction we conclude that

As, 1 mole of NaCl react with 1 mole of NH_3

So, 0.243 mole of NaCl react with 0.243 mole of NH_3

From this we conclude that, NH_3 is an excess reagent because the given moles are greater than the required moles and NaCl is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of NaHCO_3

From the reaction, we conclude that

As, 1 mole of NaCl react to give 1 mole of NaHCO_3

So, 0.243 moles of NaCl react to give 0.243 moles of NaHCO_3

Now we have to calculate the mass of NaHCO_3

$\text{ Mass of }NaHCO_3=\text{ Moles of }NaHCO_3\times \text{ Molar mass of }NaHCO_3$

Molar mass of sodium bicarbonate = 84 g/mol

$\text{ Mass of }NaHCO_3=(0.243moles)\times (84g/mole)=20.4g$

Thus, the theoretical yield of sodium bicarbonate in grams is, 20.4 grams.

Answer from: Quest

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Answer from: Quest

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.