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Answer :
Final answer:
To react completely with 39.1 mL of N2 gas at STP, approximately 0.0347 grams of lithium is required.
Explanation:
The mass of lithium required to react completely with 39.1 mL of N2 gas at STP, we need to use stoichiometry and the ideal gas law. First, we calculate the number of moles of N2 gas using the ideal gas law.
Then, we use the balanced equation of the reaction to determine the mole ratio between lithium and N2. Finally, we convert the moles of lithium to grams using its molar mass.
Given that the volume of the N2 gas is 39.1 mL, we can convert it to liters by dividing by 1000: 39.1 mL ÷ 1000 = 0.0391 L.
The ideal gas law 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 at STP (273.15 K).
Using the equation PV = nRT and substituting the values for P, V, and R, we can solve for n:
(1 atm) × (0.0391 L) = n × (0.0821 atm L/mol K) × (273.15 K).
Solving for n gives us: n = 0.00167 mol.
From the balanced equation for the reaction, we can see that 3 moles of lithium are required to react with 1 mole of N2 gas. Therefore, 0.00167 mol of N2 gas will require: 0.00167 mol N2 × (3 mol Li / 1 mol N2)
= 0.00501 mol of Li.
Finally, to convert moles of lithium to grams, we multiply by the molar mass of lithium: 0.00501 mol Li × (6.94 g/mol) = 0.0347 g.
Therefore, approximately 0.0347 grams of lithium is required to react completely with 39.1 mL of N2 gas at STP.
Learn more about stoichiometry here:
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