Technically any number you choose. I'm sure the author of the problem meant (but didn't say) how much HCl is need to react COMPLETELY with the 5.8 g Al(OH)3
3HCl + Al(OH)3 ==> AlCl3 + 3H2O
mols Al(OH)3 = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols Al(OH)3 to mols HCl. That's
?mols Al(OH)3 x (3 mols HCl/1 mol Al(OH)3) = ? mols Al(OH)3 x 3/1 = ?
Then M = mols/L.
You know M of the HCl, and mols HCl, solve for L HCl, then convert to mL.
How many mL of 1.2 M HCl is needed to react with 5.8 g of Al(OH)3
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