H2SO4(aq) + 2LiOH(aq) → Li2SO4(aq) + 2H2O(l)

Question

3.2 moles of LiOH enter the reaction and are fully consumed. The molar mass of Li2SO4 is 110 g/mol. The final aqueous solution has a volume of 4.0 L.
What is the final concentration of Li2SO4 in g/L?

44 g/L
130 g/L
440 g/L
12 g/L

I do not know how to solve this.

Answer 1

Then 3.2 mols LiOH will form 3.2/2 = 1.6 mols Li2SO4
grams Li2SO4 = mols x molar mass = ?
(Li2SO4) in g/L = grams Li2SO4/L solution.

Answer 2

That makes sense. So the answer would be 44 g/L?

Answer 3

yes

Answer 4

When aqueous lead (II) nitratre reacts with aqueous potassium iodide, a bright yellow solid is formed, lead (II) iodide. If 25.0 mL of 0.688M of potassum iodide is reacted with excess lead (II) nitrate, how many grams of lead (II) iodide would be formed? Based on the BCA table, how many moles of lead (II) iodide would be formed?

How many grams of lead (II) iodide is that?