If 35.0 mL of 0.210 M HCl reacts with excess Mg, how many mL of hydrogen gas are produced at STP?

Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)

Is the answer 2.352 STP

converted mols HCl to mols H2 gas:

02.10 M HCl x 1 mol H2/2 mol HCl = 0.105 mol H2

converted mols H2 gas to L:

0.105 mol H2 x 1L/1 mol= 0.105 L H2

multiplied L of H2 to 22.4 STP

0.105L x 22.4 STP= 2.352

No. You don't have 2.10 mols HCl. Did you read my response? mols HCl = M x L = 2.10 x 0.035 = ?

Then convert to mols H2 and from thre to L.

To find the answer to the question, you need to use stoichiometry to determine the amount of hydrogen gas produced. Here's the step-by-step explanation:

1. First, convert the volume of HCl solution (35.0 mL) to moles of HCl.
- Moles = Volume (L) x Concentration (M)
- Convert mL to L by dividing by 1000: 35.0 mL / 1000 = 0.035 L
- Moles of HCl = 0.035 L x 0.210 M = 0.00735 mol HCl

2. Next, use the balanced chemical equation to calculate the moles of hydrogen gas produced.
- According to the equation, 2 moles of HCl produce 1 mole of H2 gas.
- Moles of H2 gas = 0.00735 mol HCl x (1 mol H2 / 2 mol HCl) = 0.003675 mol H2

3. Now, convert the moles of hydrogen gas to volume at STP (Standard Temperature and Pressure).
- According to the ideal gas law, 1 mole of any gas occupies 22.4 L at STP.
- Volume of H2 gas = 0.003675 mol H2 x (22.4 L / 1 mol) = 0.08228 L H2

Therefore, the volume of hydrogen gas produced at STP is approximately 0.08228 L or 82.28 mL. The answer 2.352 STP you mentioned is not correct.