Given Zn(NO3)2 + (NH4)2S = ZnS + 2NH4NO3 (molar mass of Zn(NO3)2 189 g, molar mass of (NH4)2S 68 g), what is the molarity of a 250 ml solution if you have 123 g of zinc nitrate?
Moles zinc nitrate = grams zinc nitrate/molar mass.
M = moles/L.
I don't know what the equation has to do with the post. Please let me know if I have interpreted the problem incorrectly.
To calculate the molarity of the solution, we need to know the number of moles of Zn(NO3)2 and the volume of the solution.
First, let's calculate the number of moles of Zn(NO3)2:
Given that the molar mass of Zn(NO3)2 is 189 g/mol and we have 123 g of Zn(NO3)2, we can use the formula:
Number of moles = mass / molar mass
Number of moles of Zn(NO3)2 = 123 g / 189 g/mol
Number of moles of Zn(NO3)2 = 0.651 moles
Since the balanced equation shows that Zn(NO3)2 has a coefficient of 1, the number of moles of Zn(NO3)2 in the reaction is also equal to the number of moles of ZnS formed.
Next, let's determine the volume in liters (L) of the solution:
We have 250 mL of solution, and to calculate the molarity, we need to use liters.
1 L is equal to 1000 mL, so the volume of the solution is:
Volume of solution = 250 mL / 1000 mL/L
Volume of solution = 0.25 L
Now that we have the number of moles (0.651 moles) and the volume in liters (0.25 L), we can calculate the molarity using the formula:
Molarity (M) = Number of moles / Volume (L)
Molarity = 0.651 moles / 0.25 L
Molarity = 2.604 M
Therefore, the molarity of the 250 mL solution, given that you have 123 g of zinc nitrate, is 2.604 M.