estimate the mass of BaSO4 that will be produced when a 100ml of a 2M solution of Ba(NO3)2 in water is added toa 100ml of a 2M solution of NaSO4 in water
you havea .2Moles Ba(NO3)2
Ba(NO3)2+Na2SO4>>BaSO4 + ...
You have .2 moles of Na2SO4
you have .2 moles of Ba(NO3)2
so you must be getting .2 moles of barium sulfate.
masss=.2*molmassBaSO4
To estimate the mass of BaSO4 that will be produced, we need to use the concept of stoichiometry, which relates the number of moles of each reactant with the number of moles of the product.
First, we need to write the balanced chemical equation for the reaction between Ba(NO3)2 and Na2SO4:
Ba(NO3)2 + Na2SO4 ⟶ BaSO4 + 2 NaNO3
From the balanced equation, we can determine that one mole of Ba(NO3)2 reacts with one mole of Na2SO4 to produce one mole of BaSO4.
Next, we need to calculate the number of moles of Ba(NO3)2 and Na2SO4 in the given solutions.
We are given the concentration of both solutions as 2M, which means that there are 2 moles of Ba(NO3)2 and Na2SO4 in 1 liter (1000 mL) of the solution. Since we have 100 mL of each solution, we can calculate the number of moles in each solution using the formula:
Moles = Concentration (M) x Volume (L)
For the Ba(NO3)2 solution:
Moles of Ba(NO3)2 = 2 M x 0.1 L (100 ml = 0.1 L)
Moles of Ba(NO3)2 = 0.2 moles
For the Na2SO4 solution:
Moles of Na2SO4 = 2 M x 0.1 L (100 ml = 0.1 L)
Moles of Na2SO4 = 0.2 moles
Since the ratio in the balanced equation is 1:1, we know that the limiting reactant is the one with the smaller number of moles, which is Ba(NO3)2 in this case.
Using the stoichiometric ratio, we can say that 0.2 moles of Ba(NO3)2 will produce 0.2 moles of BaSO4.
Finally, we need to calculate the mass of BaSO4 using its molar mass. The molar mass of BaSO4 is calculated by adding the molar masses of its individual atoms:
Molar mass of BaSO4 = (molar mass of Ba) + (molar mass of S) + 4 x (molar mass of O)
Molar mass of BaSO4 = 137.3 g/mol + 32.1 g/mol + 4 x 16.0 g/mol
Molar mass of BaSO4 = 233.3 g/mol
Now, we can calculate the mass of BaSO4:
Mass of BaSO4 = moles of BaSO4 x molar mass of BaSO4
Mass of BaSO4 = 0.2 moles x 233.3 g/mol
Mass of BaSO4 = 46.66 g
Therefore, approximately 46.66 grams of BaSO4 will be produced when 100 ml of a 2M solution of Ba(NO3)2 is added to 100 ml of a 2M solution of Na2SO4.