Question

You use an analytical balance to measure the mass of a weigh boat and record a mass of 1.5624 g. You then add some NaHCO3 to the weigh boat and give the weigh boat to your lab partner to measure. Your lab partner mistakenly neglects to record all the sig figs on the balance and records a mass of 1.92 g (corrected rounded). What is the minimum number of grams of acetic acid that you would need to fully react with your NaHCO3? Report your answer with appropriate sig figs and be sure to include units.

Answer 1

Answer:

The minimum number of grams of acetic acid that we would need to fully react with your NaHCO3, is 0.263 grams.

Explanation:

Step 1: Data given

Mass weigh boat = 1.5624 grams

Mass of boat + NaHCO3 = 1.92...

Step 2: Calculate ranges

The minimum possible mass of boat + NaHCO3 = 1.9200g  

Minimum mass of NaHCO3 = 1.9200g - 1.5624g = 0.3576g  

The maximum possible mass of boat + NaHCO3 = 1.9299g  

Maximum mass of NaHCO3 = 1.9299g - 1.5624g = 0.3675g  

What is the minimum mass of acetic acid that must be used to  react with all the NaHCO3 possible in the boat:  

This means what mass of CH3COOH reacts with 0.3675g NaHCO3  

NaHCO3 + CH3COOH → CH3COONa + CO2 + H2O

1 mole of NaHCO3 consumed, needs 1 mole CH3COOH to produce 1 mole of CH3COONa, 1 mole CO2 and 1 mole H2O

Step 3: Calculate moles NaHCO3

Moles NaHCO3 = mass NaHCO3 / Molar mass NaHCO3

moles NaHCO3 = 0.3675 grams / 84g/mol

moles NaHCO3 = 0.004375 moles

Since 1 mole of NaHCO3 consumed, needs 1 mole CH3COOH

For 0.004375 moles NaHCO3, we need 0.004375 moles CH3COOH

Step 4: Calculate mass of CH3COOH

mass of CH3COOH = moles CH3COOH * Molar mass CH3COOH

mass of CH3COOH = 0.004375 moles * 60.05 g/mol

mass CH3COOH = 0.2627 grams ≈ 0.263 grams

The minimum number of grams of acetic acid that we would need to fully react with your NaHCO3, is 0.263 grams.