H2SO4(aq) + 2LiOH(aq) → Li2SO4(aq) + 2H2O(l)
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.
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.
That makes sense. So the answer would be 44 g/L?
yes
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?
To find the final concentration of Li2SO4 in g/L, we need to first find the number of moles of Li2SO4 produced and then divide it by the volume of the final solution.
From the balanced equation, we know that 1 mole of H2SO4 reacts with 2 moles of LiOH to produce 1 mole of Li2SO4. Therefore, we can say that for every 2 moles of LiOH consumed, 1 mole of Li2SO4 is produced.
Given that 3.2 moles of LiOH are consumed, this means that 3.2/2 = 1.6 moles of Li2SO4 are produced.
Next, we need to convert the moles of Li2SO4 to grams. The molar mass of Li2SO4 is given as 110 g/mol. Therefore, 1.6 moles of Li2SO4 is equal to (1.6 moles) * (110 g/mol) = 176 g of Li2SO4.
Finally, we divide the mass of Li2SO4 by the volume of the final solution. The volume is given as 4.0 L.
Therefore, the final concentration of Li2SO4 in g/L is (176 g) / (4.0 L) = 44 g/L.
Hence, the correct answer is 44 g/L.