The following reaction is carried out as a source of hydrogen gas in the laboratory:

Zn(s) + 2HCl(aq) ---> ZnCl2(aq) + H2(g)
If 305 mL of hydrogen gas is collected over water at 25 °C at a total pressure of 746 mm Hg , how many grams of Zn reacted?

See your post above.

To determine how many grams of Zn reacted, we need to use the ideal gas law equation and the stoichiometry of the balanced chemical equation.

1. Begin by converting the volume of hydrogen gas collected over water from milliliters to liters:
305 mL ÷ 1000 mL/L = 0.305 L

2. Next, convert the pressure from mm Hg to atm:
746 mm Hg ÷ 760 mm Hg/atm = 0.98 atm

3. The ideal gas law equation is given by PV = nRT, where P is pressure, V is volume, n is moles, R is the ideal gas constant, and T is temperature in Kelvin.
We need to solve for the moles of hydrogen gas:
n = (PV) / (RT)

The ideal gas constant is 0.0821 L·atm/mol·K.
The temperature should be converted from Celsius to Kelvin:
25 °C + 273 = 298 K

Plugging the values into the equation:
n = (0.98 atm * 0.305 L) / (0.0821 L·atm/mol·K * 298 K)
n ≈ 0.012 mol

4. From the balanced chemical equation, we can see that the mole ratio between Zn and H2 is 1:1.
This means that 1 mole of Zn reacts with 1 mole of H2.

Since the number of moles of hydrogen gas is equal to the number of moles of zinc reacted in the balanced chemical equation, we can conclude that 0.012 mol of Zn reacted.

5. Now we need to calculate the molar mass of Zn, which is 65.38 g/mol.

6. Finally, calculate the mass of Zn reacted:
mass = molar mass x moles of Zn
mass = 65.38 g/mol x 0.012 mol

The mass of Zn reacted is approximately 0.784 grams.