how many liters of hydrogen gas measured at 1.05 atm and negative 25 degrees will be produced when 5.00 grams of magnesium metal is dropped into excess hydrochloric acid?

To determine the number of liters of hydrogen gas produced when 5.00 grams of magnesium reacts with excess hydrochloric acid, we need to follow a few steps.

Step 1: Write and balance the chemical equation:
The balanced equation for the reaction between magnesium and hydrochloric acid is:

Mg + 2HCl → MgCl2 + H2

Step 2: Convert the mass of magnesium to moles:
Molar mass of magnesium (Mg) = 24.31 g/mol
Number of moles of Mg = mass of Mg / molar mass of Mg
= 5.00 g / 24.31 g/mol

Step 3: Use stoichiometry to find the moles of hydrogen gas:
From the balanced equation, we can see that 1 mole of magnesium (Mg) reacts to produce 1 mole of hydrogen (H2).
So, the moles of H2 = moles of Mg

Step 4: Calculate the volume of hydrogen gas using the ideal gas law:
The ideal gas law equation is:
PV = nRT

P = pressure (1.05 atm)
V = volume (unknown)
n = number of moles of gas (moles of H2)
R = ideal gas constant (0.08206 L·atm/(mol·K))
T = temperature (in Kelvin)

Given temperature is -25°C, which needs to be converted to Kelvin:
T (Kelvin) = T (°C) + 273.15
T (Kelvin) = -25°C + 273.15

Step 5: Solve for the volume (V) using the ideal gas law equation:
V = (nRT) / P

Now we can plug in the values and calculate the volume of hydrogen gas:
V = (moles of H2 * R * T) / P

Note: The given conditions state that hydrogen gas is measured at 1.05 atm and -25°C, but you haven't mentioned the volume, so we will calculate the volume in liters based on the other given values.

Please provide the number of moles of hydrogen (H2) so that we can proceed with the calculation.