how many grams of H2 (g) are produced by the reaction of 1.35 g Al with 75.0 mL of 3.44 M HCl

Here is the way to do most of these stoichiometry/limiting reagent type problems.
Step 1. Write a balanced chemical equation.
2Al + 6HCl ==? 3H2 + 2AlCl3

Step 2. Convert what you have (in this case Al metal and HCl) into mols remembering that mols = g/molar mass AND mols = L x Molarity.
a. mols Al = 1.35 g/27 (I am rounding and you will need to look up the molar masses and redo the calculations) = ?? mols Al
b. mols HCl = 0.075 L x 3.44 M = xx mols HCl

3. Convert mols of what you have (in this case Al and HCl) into mols of what you want (in this case H2) using the coefficients in the balanced equation from step 1.
a. for Al. mols H2 = ??mols Al(from above) x (3 mols H2/2 mols Al) = yy mols H2.

b. for HCl. mols H2 = xx mols HCl (from above) x (3 mols H2/6 mols HCl) = zz mols H2.

c. Obviously both answers can't be correct. Choose the smaller of a or b and proceed.

4. mols H2 from step 3a or 3b x molar mass = grams H2.

I hope this helps. Post your work if you need further assistance and tell us
where you are having trouble.

thanks! one more.

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