Magnesium nitrate reacts with lithium sulfate in a double replacement reaction. You have 3.45 g of lithium sulfate and 275 mL of magnesium nitrate. Calculate the molarity of the Magnesium nitrate solution in the reaction.
First I want to make it very clear that Mg(NO3)2 and Li2SO4 will NOT react at all and it is folly to try to calculate the M of the Mg(NO3)2 this way. However, the calculation can be done (although it doesn't man anything)so in the spirit of chemistry I will continue.
Mg(NO3)2 + Li2SO4 ==> MgSO4 + 2LiNO3
moles Li2SO4 = grams/molar mass = ?
Using the coefficients in the balanced equation, 1 mol Li2SO4 reacts with 1 mol Mg(NO3)2 so mol Li2SO4 = moles Mg(NO3)2
M Mg(NO3)2 = moles Mg(NO3)2/L Mg(NO3)2
Not possible to work, as I see it. 275mL of magnesium nitrate? How can one determine how much magnesium nitrate is there? is it in excess, or deficit, or exactly enough? One wonders.
welp that's what the problem is.
To calculate the molarity of the magnesium nitrate solution, we need to first determine the moles of magnesium nitrate present in the solution.
Step 1: Calculate the moles of lithium sulfate:
We are given that we have 3.45 g of lithium sulfate. To find the moles, we divide the given mass by the molar mass of lithium sulfate.
The molar mass of lithium sulfate (Li2SO4) is:
(2 x Atomic mass of Li) + Atomic mass of S + (4 x Atomic mass of O)
= (2 x 6.94 g/mol) + 32.07 g/mol + (4 x 16.00 g/mol)
= 109.94 g/mol
Now, we can calculate the moles of lithium sulfate:
Moles = Mass / Molar mass
Moles = 3.45 g / 109.94 g/mol
Moles = 0.0314 mol
Step 2: Calculate the moles of magnesium nitrate:
We have 275 mL of the magnesium nitrate solution. However, we need to convert this volume to moles.
To convert mL to L, we divide by 1000:
Volume (L) = 275 mL / 1000 mL/L
Volume (L) = 0.275 L
The molarity of a solution is defined as moles of solute divided by liters of solution. Since we want to calculate the molarity of magnesium nitrate, we need to find the moles.
According to the balanced chemical equation, the ratio of magnesium nitrate (Mg(NO3)2) to lithium sulfate (Li2SO4) is 1:1. This means that for every mole of lithium sulfate reacted, there is an equal number of moles of magnesium nitrate.
Therefore, the moles of magnesium nitrate in the reaction is also 0.0314 mol.
Step 3: Calculate molarity:
Molarity (M) = Moles of solute / Volume of solution (in liters)
Moles of solute (magnesium nitrate) = 0.0314 mol
Volume of solution (in liters) = 0.275 L
Molarity (M) = 0.0314 mol / 0.275 L
Molarity (M) ≈ 0.114 M
Therefore, the molarity of the magnesium nitrate solution in the reaction is approximately 0.114 M.