Respuesta :
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
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: 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:
[tex]\text{Molarity of barium nitrate}=\frac{\text{Moles of barium nitrate}}{\text{Volume of solution}}[/tex]
Putting values in above equation, we get:
[tex]0.5mol/L=\frac{\text{Moles of barium nitrate}}{0.02L}\\\\\text{Moles of barium nitrate}=0.01mol[/tex]
The equation for the chemical reaction of barium nitrate and sodium sulfate follows:
[tex]Na_2SO_4+Ba(NO_3)_2\rightarrow BaSO_4+2NaNO_3[/tex]
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 = [tex]\frac{1}{1}\times 0.01=0.01mol[/tex] of barium sulfate.
To calculate the mass of barium sulfate, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Moles of barium sulfate = 0.01 mol
Molar mass of barium sulfate = 233.38 g/mol
Putting values in above equation, we get:
[tex]0.01mol=\frac{\text{Mass of barium sulfate}}{233.38g/mol}\\\\\text{Mass of barium sulfate}=2.3338g[/tex]
Hence, the mass of barium sulfate produced will be 2.3338 grams.
