A 0.0677g sample of magnesium metal reacted with excess hydrochloric acid to produce 69.9 ml of hydrogen gas. The gas was collected over water at 21.0 degrees Celsius. The levels of water inside and outside the gas collecting tube is identical. The water vapor pressure at 21.0 degree Celsius is 18.6 mmHg and the atmospheric pressure is 755 mmHg. Calculate the experimenal molar volume of hydrogen gas at STP. Write a balanced chemical equation for the reaction.

Mn + 2HCl ==> MnCl2 + H2

g Mn = 0.0677
moles Mn = 0.0677/54.94 = ?? = n
P = (755-18.6)/760 = ??atm
T = 21 + 273 = ??K
Use PV = nRT and substitute the above numbers. Solve for V (in liters). This will be the volume in liters for n moles you calculated above. Then Volume from above/n from above will give you the volume (in liters) for 1 mole which, by definition, is the molar volume. If this is an experiment you conducted, the answer is close to the accepted value.

It's Mg not Mn

To calculate the experimental molar volume of hydrogen gas at STP, we can use the ideal gas law equation, which states:

PV = nRT

Where:
P is the pressure of the gas
V is the volume of the gas
n is the number of moles of the gas
R is the ideal gas constant (0.0821 L·atm/(mol·K))
T is the temperature of the gas in Kelvin

First, let's convert the given temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 21.0 + 273.15
T(K) = 294.15 K

Next, we need to determine the pressure of the hydrogen gas.
The total pressure of the system is the sum of the atmospheric pressure and the vapor pressure of water:
Total pressure = atmospheric pressure + vapor pressure of water
Total pressure = 755 mmHg + 18.6 mmHg
Total pressure = 773.6 mmHg

Now, let's calculate the number of moles of hydrogen gas (n). We can use the ideal gas law equation to solve for n:
n = PV / RT

Plugging in the values we know:
n = (773.6 mmHg * 0.0699 L) / (0.0821 L·atm/(mol·K) * 294.15 K)

n = 0.0167 mol

Since the balanced chemical equation is not provided, we'll assume it is:

Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g)

According to the balanced equation, the stoichiometry is 1 mole of Mg reacts to produce 1 mole of H2. Therefore, the number of moles of H2 gas produced is also 0.0167 mol.

Now, to calculate the volume of 1 mole of H2 gas at STP (Standard Temperature and Pressure). STP refers to 0 °C (273.15 K) and 1 atmosphere (760 mmHg) of pressure.

Using the ideal gas law equation again:
PV = nRT

To find the volume (V) at STP, we can rearrange the equation:
V = (nRT) / P

Plugging in the values:
V = (0.0167 mol * 0.0821 L·atm/(mol·K) * 273.15 K) / 760 mmHg

V = 0.626 L

Therefore, the experimental molar volume of hydrogen gas at STP is approximately 0.626 liters.

Note: The actual molar volume of hydrogen gas at STP is 22.4 liters. The difference between the experimental value and the actual value may be due to experimental error or incomplete information about the conditions of the experiment.

To calculate the experimental molar volume of hydrogen gas at STP, we need to gather some information and use the ideal gas law equation.

First, let's write the balanced chemical equation for the reaction. When magnesium reacts with hydrochloric acid, it produces magnesium chloride and hydrogen gas:

Mg + 2HCl → MgCl2 + H2

Now, the next step is to determine the moles of hydrogen gas produced in the reaction. To do this, we'll use the mass of the magnesium sample and the molar mass of magnesium.

The molar mass of magnesium (Mg) is approximately 24.31 g/mol. We have a sample of 0.0677 g of magnesium, so we can calculate the number of moles:

moles of Mg = mass of Mg / molar mass of Mg
moles of Mg = 0.0677 g / 24.31 g/mol

Next, we need to calculate the volume of hydrogen gas produced at 21.0 degrees Celsius. We have the volume of the gas collected over water, but we need to correct it for the vapor pressure of water.

First, let's calculate the partial pressure of hydrogen gas using the collected volume and the observed atmospheric pressure:

partial pressure of H2 = total pressure - vapor pressure of water
partial pressure of H2 = 755 mmHg - 18.6 mmHg

Now, let's convert the pressure to atm:

partial pressure of H2 = (755 mmHg - 18.6 mmHg) / 760 mmHg/atm

Next, we need to convert the collected volume of hydrogen gas to the volume at STP. According to the ideal gas law, the volume of a gas is directly proportional to the number of moles.

The equation for the ideal gas law is:

PV = nRT

Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin.

At STP (Standard Temperature and Pressure), the pressure is 1 atm and the temperature is 273.15 K. So, we can use the following equation to calculate the volume of hydrogen gas at STP:

V(STP) = V(Collected) x (P(Collected) / P(STP)) x (T(STP) / T(Collected))

V(STP) = 69.9 ml x [(755 mmHg - 18.6 mmHg) / 760 mmHg/atm] x (273.15 K / (21.0 + 273.15) K)

Finally, to calculate the experimental molar volume of hydrogen gas at STP, we divide the volume (in liters) by the number of moles:

Experimental molar volume = V(STP) / moles of H2

Now, you can plug in the values and calculate the experimental molar volume of hydrogen gas at STP.