Question

If an enclosure of 0.462 L has a partial pressure of O2 of 3.5×10⁻⁶ torr at 26 °C, what mass of magnesium will react according to the following equation? 2Mg(s) + O2(g) → 2MgO(s). Express your answer using two significant figures.

A) 0.01 grams
B) 0.02 grams
C) 0.03 grams
D) 0.04 grams

Answer 1

To calculate the mass of magnesium that will react with the given amount of oxygen, we can use the ideal gas law to determine the number of moles of oxygen. By multiplying the number of moles of oxygen by the molar mass of magnesium, we can find the mass of magnesium that will react. In this case, the mass is 0.23 grams. The correct answer is (C) 0.03 grams.

Step-by-step explanation:

To calculate the mass of magnesium that will react with the given amount of oxygen, we need to determine the number of moles of oxygen first. From the partial pressure of O2, we can calculate the number of moles using the Ideal Gas Law. The equation is 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.

Given: P = 3.5x10^-6 torr, V = 0.462 L, R = 0.0821 L·atm/mol·K, T = 26°C = 26 + 273 = 299 K

Solving the Ideal Gas Law equation for n, we get n = PV / RT. Substituting the given values, we have n = (3.5x10^-6 torr)(0.462 L) / (0.0821 L·atm/mol·K)(299 K)

n = 5.84x10^-9 mol

According to the balanced equation, 2 moles of magnesium react with 1 mole of oxygen to produce 2 moles of magnesium oxide. Therefore, the mass of magnesium that will react is 2 times the number of moles of oxygen.

Mass of magnesium = 2(5.84x10^-9 mol)(24.31 g/mol)

Mass of magnesium = 2.27x10^-7 g

Rounded to two significant figures, the mass of magnesium that will react is 0.23 grams. Therefore, the correct answer is (C) 0.03 grams.