Zinc reacts with 387 mL of 4.00 M cold, aque-

ous sulfuric acid through single replacement.
a. How much zinc sulfate is produced?
Answer in units of g
b. How many liters of hydrogen gas would be
released at STP?
Answer in units of L

-- I have solved part a, but I am confused on part B. I would think that since there is only 1 mol of hydrogen gas produced, that at STP, there would be 22.4 L of H2 produced, but I know that is the incorrect answer. I would know how to solve B if I knew that the original information was also at STP. Can you assume that it is?

I would assume that since an amount for Zn is not given that H2SO4 is the limiting reagent (LR).

Zn + H2SO4 ==> ZnSO4 + H2
You have 4.00 x 0.387 = about 1.55 mols H2SO4 which will produce 1.55 mols H2 at STP. (Note that you don't have 1 mol H2 which is the error in your thinking.)
Since 1 mol of a gas at STP occupies 22.4L, 1.55 mols will occupy 22.4 x 1.55 = ? Remember 1.55 is an estimate; you calculate should confirm that.

To solve part B, we need to use the stoichiometry of the reaction to determine the number of moles of hydrogen gas produced, and then convert it to liters at STP.

First, let's write the balanced chemical equation for the reaction between zinc and sulfuric acid:

Zn + H₂SO₄ → ZnSO₄ + H₂

According to the balanced equation, one mole of zinc reacts with one mole of sulfuric acid, producing one mole of zinc sulfate and one mole of hydrogen gas.

Since we know the volume and concentration of the sulfuric acid, we can calculate the number of moles of sulfuric acid:

Moles of H₂SO₄ = volume (in liters) × concentration (in mol/L)

Given:
Volume of H₂SO₄ = 387 mL = 387/1000 = 0.387 L (convert mL to L)
Concentration of H₂SO₄ = 4.00 M

Moles of H₂SO₄ = 0.387 L × 4.00 mol/L = 1.548 mol

Since the reaction is a 1:1 ratio between H₂SO₄ and H₂, we can conclude that 1.548 moles of hydrogen gas are produced.

Now, to convert moles of hydrogen gas to liters at STP, we can use the ideal gas law:

PV = nRT

At STP (standard temperature and pressure), T = 273.15 K and P = 1 atm.

We can rearrange the ideal gas law equation to solve for the volume:

V = nRT / P

Substituting the values:
n = 1.548 mol
R = 0.0821 L·atm/(mol·K)
T = 273.15 K
P = 1 atm

V = (1.548 mol) × (0.0821 L·atm/(mol·K)) × (273.15 K) / (1 atm)
V = 36.3 L

Therefore, the correct answer is 36.3 liters (L) of hydrogen gas would be released at STP.