Aqueous sodium sulfate reacts with aqueous barium chloride to produce solid barium sulfate and aqueous sodium chloride. the solution contains 3.44g of barium chloride. you have a 0.282M solution of sodium sulfate. determine the volume of the sodium sulfate solution necessary to react with all of the barium chloride
BaCl2 + Na2SO4 => 2NaCl + BaSO4
Moles BaCl2 = Moles Na2SO4
Moles = Molarity x Volume(liters)
Moles BaCl2 = (3.44gms/207g/mol)= 0.0166 mole BaCl2 which will use the same number of moles Na2SO4.
Therefore, Moles Na2SO4 used = 0.0166 mole = (0.282M Na2SO4)(Volume in Liters) => Vol(L) = 0.059L = 59-ml of the 0.282M Na2SO4 solution.
Oh, chemistry! You sure know how to make things interesting. Alright, let's see what we can do here.
First, we need to figure out how many moles of barium chloride we have. We can do that by using the formula:
moles = mass / molar mass
The molar mass of barium chloride (BaCl2) is 208.23 g/mol. So, plugging in the values:
moles = 3.44g / 208.23 g/mol
Calculating that gives us approximately 0.0165 moles of barium chloride.
Now, let's check out the balanced equation for the reaction:
Na2SO4 + BaCl2 ⟶ BaSO4 + 2NaCl
From the equation, we can see that the ratio of barium chloride to sodium sulfate is 1:1. So, we need an equal number of moles of sodium sulfate to react with all the barium chloride.
Since we have a 0.282M (moles per liter) solution of sodium sulfate, we can use the formula:
moles = Molarity × volume (in liters)
Rearranging that formula, we get:
volume (in liters) = moles / Molarity
Plugging in the values:
volume (in liters) = 0.0165 moles / 0.282 mol/L
Calculating that gives us approximately 0.0585 liters, or 58.5 milliliters.
So, the volume of the sodium sulfate solution necessary to react with all of the barium chloride is approximately 58.5 mL.
Remember, though, that chemistry can be a bit unpredictable sometimes, just like my jokes. So make sure to double-check your calculations and always wear safety goggles!
To determine the volume of the sodium sulfate solution necessary to react with all of the barium chloride, we need to use the stoichiometry of the reaction.
The balanced chemical equation for the reaction is:
Na2SO4(aq) + BaCl2(aq) → BaSO4(s) + 2NaCl(aq)
From the equation, we can see that each 1 mole of barium chloride reacts with 1 mole of sodium sulfate.
First, we calculate the number of moles of barium chloride using its molar mass:
Molar mass of BaCl2 = atomic mass of Ba + 2 * atomic mass of Cl = 137.33 g/mol + 2 * 35.45 g/mol = 208.23 g/mol
Number of moles of BaCl2 = 3.44 g / 208.23 g/mol = 0.0165 mol
Since the reaction is 1:1 between barium chloride and sodium sulfate, we need 0.0165 moles of sodium sulfate.
Now let's calculate the volume of the sodium sulfate solution using the molarity and the number of moles:
Molarity of Na2SO4 solution = 0.282 mol/L
Number of moles required = 0.0165 mol
Volume of sodium sulfate solution = (0.0165 mol) / (0.282 mol/L) = 0.0585 L
Finally, we convert the volume to milliliters (mL) by multiplying by 1000:
Volume of sodium sulfate solution = 0.0585 L * 1000 mL/L = 58.5 mL
Therefore, you need 58.5 mL of the 0.282M sodium sulfate solution to react with all of the 3.44g of barium chloride.
To determine the volume of the sodium sulfate solution necessary to react with all of the barium chloride, we need to use stoichiometry to calculate the number of moles of barium chloride and then use the molarity of sodium sulfate to find the volume.
First, let's calculate the number of moles of barium chloride:
Mass of barium chloride = 3.44g
To convert grams to moles, we need the molar mass of barium chloride. The molar mass of BaCl2 is:
Ba: atomic mass = 137.33 g/mol
Cl: atomic mass = 2 * 35.45 g/mol = 70.9 g/mol
Total molar mass of BaCl2 = 137.33 g/mol + 70.9 g/mol = 208.23 g/mol
Number of moles of BaCl2 = Mass of BaCl2 / Molar mass of BaCl2
Number of moles of BaCl2 = 3.44g / 208.23 g/mol
Now, let's calculate the volume of the sodium sulfate solution using the molarity:
Molarity (M) = Moles of solute / Volume of solution (in liters)
We know the molarity of the sodium sulfate solution is 0.282M and we have calculated the number of moles of BaCl2. Rearrange the equation to solve for the volume of solution:
Volume of solution (in liters) = Moles of solute / Molarity
Volume of solution (in liters) = (3.44g / 208.23 g/mol) / 0.282 mol/L
Finally, plug in the values and calculate the volume:
Volume of solution (in liters) = (0.016 Boyles) / (0.282 mol/L)
Thus, the volume of the sodium sulfate solution necessary to react with all of the barium chloride is 0.057 L (or 57 mL).