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Question 8.CC.3: What is the volume, in liters, of N2 required to react with ......

What is the volume, in liters, of N_2 required to react with 18.5 g of magnesium at a pressure of 1.20 atm and a temperature of 303 K?

3 Mg (s)+ N _2(g) \longrightarrow Mg _3 N _2(s)
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Initially, we convert the grams of Mg to moles and use a mole–mole factor from the balanced equation to calculate the moles of N_2 gas.

18.5\cancel{\ g\ Mg }\times \frac{1\cancel{ \text { mole } Mg} }{24.31\cancel{\ g\ Mg }} \times \frac{1 \text { mole }N_2}{3\cancel{ \text { moles } Mg} }=0.254 \text { mole of } N _2

Now, we use the moles of N_2 in the ideal gas law equation and solve for liters, the needed quantity.

V=\frac{n R T}{P}=\frac{0.254 \cancel{\text { mole } N _2} \times \frac{0.0821\ L\ \cdot\ \cancel{atm} }{\cancel{\text { mole }} \cdot\ \not K } \times 303\ \cancel K }{1.20\ \cancel{atm} }=5.27\ L

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