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 0.8176 L
Convert moles of HCl:
2.10 x 0.035= 0.00735 mol of HCl
Convert to moles of H2:
0.0735 x 1 mol H2/ 2 mols HCl= 0.03675 mol of H2
Convert to STP (L):
0.0365mol H2 x 22.4 L/1mol H2=0.8176
I think your last step is a little off; I have 82.3 mL.
I agree with your numbers up to multiplying by 22.4.
To solve this problem, you need to use stoichiometry to convert the given amount of HCl to the amount of hydrogen gas produced. Here are the steps to follow:
1. Calculate the moles of HCl present:
Given that the volume of HCl is 35.0 mL and the concentration is 0.210 M, you can calculate the moles of HCl using the following formula:
moles of HCl = volume (L) x concentration (M)
Convert 35.0 mL to L: 35.0 mL / 1000 mL/L = 0.035 L
moles of HCl = 0.035 L x 0.210 M = 0.00735 mol
2. Use the stoichiometry of the balanced equation to relate the moles of HCl to moles of hydrogen gas:
From the balanced equation, you can see that 2 moles of HCl react to produce 1 mole of H2.
moles of H2 = moles of HCl x (1 mol H2 / 2 mol HCl) = 0.00735 mol x (1 mol H2 / 2 mol HCl) = 0.003675 mol
3. Convert the moles of hydrogen gas to volume at STP:
To convert the moles of H2 to volume at STP, you need to use the molar volume of a gas at STP, which is 22.4 L/mol.
volume of H2 = moles of H2 x (22.4 L/mol) = 0.003675 mol x (22.4 L/mol) = 0.082176 L
Therefore, the volume of hydrogen gas produced at STP is 0.082176 L, which is equivalent to 0.8176 L (rounded to four decimal places).
So, the correct answer to the question "How many mL of hydrogen gas are produced at STP?" is 0.8176 L.