Ca + 2HCl --> CaCl2 + H2
mols Ca = grams/atomic mass
mols HCl = twice that since 1 mol Ca = 2 mols HCl
Then M HCl = mols HCl/L HCl and this can be arranged to L HCl = mols HCl/M HCl which let's you solve for L HCl.
What volume of 1.34 mol/L hydrochloric acid, HCI, is expected to react completely with 6.89 g of calcium?
1 answer