4 gram of Na metal introduce into a 4 litre flask filled with Cl gas at STP(273K, 1 bar) after reaction over , find limiting reactant
To find the limiting reactant in a chemical reaction, we need to compare the stoichiometry of the reactants with their respective amounts.
In this case, we have 4 grams of Na metal and Cl gas filled in a 4-liter flask at STP. The balanced chemical equation for the reaction between Na metal and Cl gas is:
2Na + Cl2 -> 2NaCl
From the equation, it is clear that the stoichiometric ratio between Na and Cl2 is 2:1. This means that for every 2 moles of Na, we need 1 mole of Cl2 to completely react.
To calculate the number of moles of Na, we can use the molar mass of Na, which is 22.99 g/mol. Therefore, 4 grams of Na is equal to:
4 g Na * (1 mol Na / 22.99 g Na) = 0.174 mol Na
To calculate the number of moles of Cl2, we can use the Ideal Gas Law at STP, where the pressure is 1 bar and the volume is 4 liters. 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 gas constant (0.0821 L∙atm/(mol∙K)), and T is the temperature in Kelvin.
By rearranging the equation, we can solve for the number of moles:
n = PV / RT
n = (1 bar) * (4 L) / (0.0821 L∙atm/(mol∙K)) * (273 K)
n ≈ 0.196 mol Cl2
Now we can compare the stoichiometric ratio with the actual ratio of moles. The ratio of moles of Na to Cl2 is:
0.174 mol Na / 0.196 mol Cl2 ≈ 0.889
Since the stoichiometric ratio is 2:1 (or 2), and the actual ratio is 0.889, we can see that Na is the limiting reactant.