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.
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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.
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
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
1 answer