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Question: Magnesium can be used as a “getter” in evacuated enclosures to react with the last traces of oxygen....

Magnesium can be used as a “getter” in evacuated enclosures to react with the last traces of oxygen. (The magnesium is usually heated by passing an electric current through a wire or ribbon of the metal.) If an enclosure of 0.452 L has a partial pressure of O2{O_2} of 3.5×10−63.5 \times {10^{ - 6}} torr at 27oC{27^o}C, what mass of magnesium will react according to the following equation?
2Mg(s)+O2(g)→2MgO(s)2Mg(s) + {O_2}(g) \to 2MgO(s)

Explanation

Solution

An ideal gas is defined as a hypothetical gas whose behaviour is independent of attractive and repulsive forces that are observed in real gases and can be completely described by the ideal gas law. The ideal gas law is derived from Charles Law, Boyle’s law and Avogadro’s law. In actuality, there is no such thing as an ideal gas, but an ideal gas is useful in easing the calculations when dealing with gases.

Complete answer:
We know that according to the ideal gas equation:
⇒PV=nRT\Rightarrow PV = nRT
We are given that the volume of the vessel was 0.452 Litres. We are also given that the partial pressure of oxygen is 3.5×10−63.5 \times {10^{ - 6}} torr which is equal to 4.6×10−9atm4.6 \times {10^{ - 9}}atm
The temperature is given to us as 27 degree Celsius or 300 K
The value of R is given as 0.082L.atm/K.mol0.082L.atm/K.mol
From this data we can calculate the number of moles of oxygen in the given reaction
⇒nO2=P×VR×T\Rightarrow {n_{{O_2}}} = \dfrac{{P \times V}}{{R \times T}}
⇒nO2=4.6×10−9atm×0.452 L0.082L.atm/K.mol×300K\Rightarrow {n_{{O_2}}} = \dfrac{{4.6 \times {{10}^{ - 9}}atm \times 0.452{\text{ }}L}}{{0.082L.atm/K.mol \times 300K}}
⇒nO2=8.5×10−11moles\Rightarrow {n_{{O_2}}} = 8.5 \times {10^{ - 11}}moles
From the equation we know that 2 moles of magnesium react with 1 mole of oxygen
So we can say number of moles of magnesium reacting here will be 2 times that of oxygen
⇒2×nO2=2×8.5×10−11moles\Rightarrow 2 \times {n_{{O_2}}} = 2 \times 8.5 \times {10^{ - 11}}moles
⇒nMg=1.7×10−10moles\Rightarrow {n_{Mg}} = 1.7 \times {10^{ - 10}}moles
To calculate the mass of magnesium reacting
Mass= Number of moles ×\times molar mass
⇒mMg=1.7×10−10moles×24.305g/mol\Rightarrow {m_{Mg}} = 1.7 \times {10^{ - 10}}moles \times 24.305g/mol
⇒mMg=4.1×10−9g\Rightarrow {m_{Mg}} = 4.1 \times {10^{ - 9}}g
The mass of magnesium that will react according to the equation is 4.1×10−9g4.1 \times {10^{ - 9}}g.

Note:
Always take care while using the value of Universal gas constant. The value depends on the unit the other quantities are in. If the terms are all in joules, the value of R is 8.314. But if the terms are in the Litre atmosphere then use R as 0.08206.
We can also use the ideal gas equation to calculate the effect of changes in any of the conditions on any of the other parameters.