Final answer:
To calculate the volume of hydrogen gas produced, use the Ideal Gas Law equation PV = nRT and the stoichiometry of the balanced chemical equation. By calculating the number of moles of H2 produced, converting the given pressure and temperature to the appropriate units, and plugging in the values into the equation, you can find the volume of H2 gas produced. The volume is 0.043 L, which is equivalent to 43 mL.
Step-by-step explanation:
To calculate the volume of hydrogen gas produced, we need to use the Ideal Gas Law equation PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. Rearranging the equation to solve for V, we get V = (nRT) / P.
First, we need to calculate the number of moles of hydrogen gas produced. We can use the stoichiometry of the balanced chemical equation to determine the ratio between magnesium (Mg) and hydrogen (H2). The balanced chemical equation is 2 Mg + 2 HCl -> 2 MgCl2 + H2, which tells us that 2 moles of Mg produce 1 mole of H2. So, the number of moles of H2 produced can be calculated as follows:
Moles of H2 = (moles of Mg / 2) = (13.5 g / 24.305 g/mol) / 2 = 0.278 mol.
Now, we can plug in the values into the equation V = (nRT) / P. R is the ideal gas constant, which is 0.0821 L·atm/(mol·K) for this calculation. We convert the given pressure of 837 mmHg to atmospheres by dividing by 760 mmHg/atm. The temperature needs to be in Kelvin, so we add 273.15 to 22 °C to get 295.15 K. The calculation is as follows:
V = (0.278 mol * 0.0821 L·atm/(mol·K) * 295.15 K) / (837 mmHg / 760 mmHg/atm) = 0.043 L, which is equivalent to 43 mL.
The volume of H2 gas produced when 13.5 g of Mg reacts is approximately 2.26 liters at standard conditions of 22 °C and 837 mmHg.
To calculate the volume of H2 gas produced from the reaction of 13.5 g of Mg with HCl, use the molar mass of Mg (24.305 g/mol) to convert the mass of Mg to moles. According to the stoichiometry of the balanced chemical equation Mg + 2 HCl → MgCl2 + H2, 1 mole of Mg produces 1 mole of H2. Once we have the moles of H2, we can use the ideal gas law (PV = nRT) to calculate the volume of H2 at the given conditions of temperature (22 °C) and pressure (837 mmHg).
Conversion from Mass of Mg to Moles
13.5 g Mg × (1 mol Mg / 24.305 g Mg) = 0.555 moles of Mg
0.555 moles of Mg will produce 0.555 moles of H2 based on the stoichiometry of the reaction.
Calculating the Volume of H2
First convert the pressure from mmHg to atm: 837 mmHg / 760 mmHg per atm = 1.10 atm. Then convert the temperature from degrees Celsius to Kelvin: 22 °C + 273 = 295 K. Now apply the ideal gas law:
PV = nRT, where P is pressure in atm, V is volume in L, n is moles of gas, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is temperature in K.
(1.10 atm) · V = (0.555 moles) · (0.0821 L·atm/mol·K) · (295 K)
V = (0.555 moles · 0.0821 L·atm/mol·K · 295 K) / 1.10 atm
V = 12.86 L·K·mol-1 · K / atm
V ≈ 2.26 L
Therefore, the correct answer is (b) 2.26 L.