This is a limiting reagent (LR) problem. You know that afbecasue amounts are given for BOTH reactants. Here are the steps. Print this out; it will work all of your LR problems.
1. Write and balance the equation. You've done that.
2. Convert grams to mols
mols = grams/molar mass. Do that for Al(OH)3 AND for H2SO4
3. Using the coefficients in the balanced equation, convert mols Al(OH)3 to mols Al2(SO4)3
4. Do the same and convert mols H2SO4 to mols Al2(SO4)3.
5. It is likely that the mols Al2(SO4)3 will NOT be the same. The correct value in LR problems is ALWAYS the smaller of the two and the reagent responsible for that smaller number is the LR.
6. Now convert that value into grams. g = mols x molar mass = ?
Post your work if you get stuck
PLEASE HELP
how many grams of AL2(SO4)3 can be made by reacting 50ml 2.8M H2SO4 with 7.4g of AL(OH)3?
3 H2SO4 + 2 Al (OH)3 ----> 6 H20 + AL2(SO4)3
4 answers
Thank you so much for your help!
One question,
For the 50ml 2.8M H2SO4 conversion, how do i convert it? Do i go for 2.8M to molar mass or 50ml to molar mass? and what do i do with the other unit?
One question,
For the 50ml 2.8M H2SO4 conversion, how do i convert it? Do i go for 2.8M to molar mass or 50ml to molar mass? and what do i do with the other unit?
and do i do step 2 and 3 in one line? like the molar mass conversion and the coefficient conversion?
You can do it all in one line but I think it's easier to do it in two in order not to be confused. To convert 50 mL of 2.8 M H2SO4 to mols (sorry I forgot that ) you do
0.05 x 2.8 = ?mols H2SO4.
0.05 x 2.8 = ?mols H2SO4.
1 answer