Question

Barium nitrate reacts with aqueous sodium sulfate to produce solid barium sulfate and aqueous sodium nitrate. abigail places 20.00 ml of 0.500 m barium nitrate in a flask. how many grams of barium sulfate will be produced?

Answer 1

I will assume the chemical reaction equation is like this:
Ba(NO3)2 + Na2SO4 ---> 2 NaNO3 + BaSO4

For every 1 barium nitrate molecule used, there will be 1 barium sulfate formed. The number of molecule barium sulfate formed would be: 20.00 ml * 0.500 mol/1000ml * 1= 0.01 mol

The mass of barium sulfate produced: 0.01mol * 233.38 g/mol= 2.3338 grams

Answer 2

Answer: The mass of barium sulfate produced will be 2.3338 grams.

Explanation:

We are given:

Molarity of barium nitrate = 0.5 M

Volume of solution = 20 mL = 0.02 L   (Conversion factor: 1 L = 1000 mL)

To calculate the number of moles of barium nitrate, we use the equation:

$\text{Molarity of barium nitrate}=\frac{\text{Moles of barium nitrate}}{\text{Volume of solution}}$

Putting values in above equation, we get:

$0.5mol/L=\frac{\text{Moles of barium nitrate}}{0.02L}\\\\\text{Moles of barium nitrate}=0.01mol$

The equation for the chemical reaction of barium nitrate and sodium sulfate follows:

$Na_2SO_4+Ba(NO_3)_2\rightarrow BaSO_4+2NaNO_3$

By Stoichiometry of the reaction:

1 mole of barium nitrate produces 1 mole of barium sulfate.

So, 0.01 moles of barium nitrate will produce = $\frac{1}{1}\times 0.01=0.01mol$ of barium sulfate.

To calculate the mass of barium sulfate, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of barium sulfate = 0.01 mol

Molar mass of barium sulfate = 233.38 g/mol

Putting values in above equation, we get:

$0.01mol=\frac{\text{Mass of barium sulfate}}{233.38g/mol}\\\\\text{Mass of barium sulfate}=2.3338g$

Hence, the mass of barium sulfate produced will be 2.3338 grams.