Zn+ 2HCl -> ZnCL2+h2

what volume of hydrogen at stp is produced when 2.5 of zinc react with an excess of hydrochloric acid

Balance the equation.

Convert 2.5 g Zn to mols. mols = g/molar mass.

Using the coefficients in the balanced equation, convert mols Zn to mols H2.

Convert mols H2 to volume H2 remembering that mols x 22.4 L/mol at stp = L.

Post your work if you get stuck.

Is it 0.86L of H2

I did
2.5 divided by 65.39 multiplied by 22.4

To determine the volume of hydrogen gas produced when 2.5 grams of zinc (Zn) reacts with an excess of hydrochloric acid (HCl) at STP (Standard Temperature and Pressure), we need to use the ideal gas law equation:

PV = nRT

Where:
P = Pressure (in atm)
V = Volume (in liters)
n = Number of moles
R = Ideal gas constant (0.0821 L·atm/mol·K)
T = Temperature (in Kelvin)

First, we need to calculate the number of moles of zinc (n) using its molar mass, which is 65.38 g/mol:
n = Mass / Molar mass
n = 2.5 g / 65.38 g/mol
n ≈ 0.0382 mol

According to the balanced chemical equation, for every 1 mole of zinc, 1 mole of hydrogen gas is produced. Therefore, the number of moles of hydrogen gas is also 0.0382 mol.

Next, we convert the temperature to Kelvin. Since STP is 273.15 K and 1 atm, we have:
T = 273.15 K
P = 1 atm

Now we can solve for the volume of hydrogen gas:
PV = nRT
V = (nRT) / P
V = (0.0382 mol) * (0.0821 L·atm/mol·K) * (273.15 K) / (1 atm)
V ≈ 0.850 L

Therefore, approximately 0.850 liters of hydrogen gas are produced at STP when 2.5 grams of zinc react with an excess of hydrochloric acid.

586.26 L of H2