Question

If 10.0 grams of NaHCO3 is added to 10.0 g of HCl, determine the efficiency of baking soda as an antacid if 6.73 g of NaCl was produced from the reaction. (In other words, calculate the percent yield of sodium chloride produced).

Answer 1

Answer:

The percent yield of this reaction is 96.8 %

Explanation:

Step 1: Data given

Mass of NaHCO3 = 10.0 grams

Mass of HCl = 10.0 grams

MAss of NaCl produced = 6.73 grams

Molar mass of NaHCO3 = 84.0 g/mol

Molar mass HCl = 36.46 g/mol

Molar mass NaCl = 58.44 g/mol

Step 2: The balanced equation

NaHCO3 (aq) + HCl (aq) → NaCl (aq) + CO2 (g) + H2O (l)

Step 3: Calculate moles

Moles = mass / molar mass

Moles NaHCO3 = 10.0 grams / 84.0 g/mol

Moles NaHCO3 = 0.119 moles

Moles HCl = 10.0 / 36.46 g/mol

Moles HCl = 0.274 moles

Step4: Calculate limiting reactant

For 1 mol NaHCO3 we need 1 mol HCl to produce 1 mol NaCl, 1 mol CO2 and 1 mol H2O

NaHCO3 is the limiting reactant. It will completely be consumed (0.119 moles). HCl is in excess. There will react 0.119 moles. There will remain 0.274 - 0.119 = 0.155 moles

Step 5: Calculate moles NaCl

For 1 mol NaHCO3 we need 1 mol HCl to produce 1 mol NaCl, 1 mol CO2 and 1 mol H2O

For 0.119 moles NaHCO3 we'll have 0.119 moles NaCl

Step 6: Calculate mass NaCl

Mass NaCl = moles NaCl * molar mass NaCl

Mass NaCl = 0.119 moles NaCl * 58.44 g/mol

Mass NaCl = 6.95 grams

Step 7: Calculate the percent yield

Percent yield =(actual mass / theoretical mass) * 100%

Percent yield = (6.73 grams / 6.95 grams) * 100%

PErcent yield = 96.8 %

The percent yield of this reaction is 96.8 %

Answer 2

Answer:

percentage yield of NaCl = 96.64%

Explanation:

The reaction was between NaHCO3 and HCl .The chemical equation can be represented below:

NaHCO3 + HCl → NaCl + H2O + CO2 . The balance equation is

NaHCO3 + HCl → NaCl + H2O + CO2

The question ask us to calculate the percentage yield of NaCl.

The efficiency of NaHCO3 as an antacid , the limiting reactant is NaHCO3

as

1 mole of NaHCO3 produces 1 mole of NaCl

Therefore,

molar mass of NaHCO3 = 23 +1 + 12 + 48 = 84 g

molar mass of NaCl = 23 + 35.5 = 58.5 g

1 mole of NaHCO3 = 84 g

1 mole of NaCl  = 58.5 g

since 84 g of NaHCO3 produces 58.5 g of NaCl

10 g of NaHCO3 will produce ? grams of NaCl

cross multiply

Theoretical yield of NaCl = (10 × 58.5)/84

Theoretical yield of NaCl = 585/84

Theoretical yield of NaCl  = 6.9642857143 g

percentage yield of NaCl = actual yield/theoretical yield × 100

percentage yield of NaCl = 6.73/6.9642857143 × 100

percentage yield of NaCl = 673/6.9642857143

percentage yield of NaCl = 96.635897436%

percentage yield of NaCl = 96.64%