Heating 1.124 g of an unknown NaHCO3-NaCl mixture resulted in a mass loss of 0.270 g.

a. Calculate the mass of NaHCO3 present in the original mixture.

b. Calculate the mass percent of NaHCO3 in the unknown mixture to the nearest
percent.

a) 0.731 grams

b) 65.0 %

Explanation:

Step 1: Data given

Mass of unknown NaHCO3- NaCl = 1.124 grams

Mass loss = 0.270 grams

Molar mass NaHCO3 = 84.00 g/mol

Step 2: The balanced equation

2 NaHCO3 → Na2CO3 + H2O + CO2

Step 3:  Calculate moles H2CO3

All the mass lost was due to an equimolar mixture of H2O and CO2, effectively H2CO3.

Moles H2CO3 = mass H2CO3 / molar mass H2CO3

Moles H2CO3 = 0.270 grams / 62.03 g/mol

Moles H2CO3 = 0.00435 moles

Step 4: Calculate moles NaHCO3

For 1 mole H2CO3 produced we need 2 moles NaHCO3

For 0.00435 moles H2CO3 we need 2*0.00435 = 0.00870 moles NaHCO3

Step 5: Calculate mass NaHCO3

Mass NaHCO3 = moles * molar mass

Mass NaHCO3 = 0.00870 moles * 84.00 g/mol

Mass NaHCO3 = 0.731 grams

b. Calculate the mass percent of NaHCO3 in the unknown mixture to the nearest  percent.

mass % = (0.731/1.124) *100 %

mass % = 65.0 %

the answer is definitely c.