Answer:
75.9 grams of salt
Explanation:
The reaction is the following:
2Na(s) + Cl₂(g) → 2NaCl(s) (1)
We have:
m(Na): the mass of sodium = 30 g
V(Cl₂): the volume of the chlorine gas at STP = 60 L
So, to find the mass of NaCl we need to calculate the number of moles of Na and Cl₂.
[tex] n_{Na} = \frac{m}{A_{r}} = \frac{30 g}{22.99 g/mol} = 1.30 moles [/tex]
The number of moles of Cl₂ can be found by the Ideal gas law equation:
[tex] PV = n_{Cl_{2}}RT [/tex]
Where:
P: is the pressure = 1 atm (at STP)
R: is the gas constant = 0.082 L*atm/(K*mol)
T: is the temperature = 273 K (at STP)
[tex] n_{Cl_{2}} = \frac{PV}{RT} = \frac{1 atm*60 L}{0.082 L*atm/(K*mol)*273 K} = 2.68 moles [/tex]
Now we need to find the limiting reactant. From the stoichiometric relation between Na and Cl₂ (equation 1), we have that 2 moles of Na react with 1 mol of Cl₂, so:
[tex] n_{Na} = \frac{2 moles Na}{1 mol Cl_{2}}*2.68 moles Cl_{2} = 5.36 moles [/tex]
Since we have 1.30 moles of Na, the limiting reactant is Na.
Finally, we can find the number of moles of NaCl and its mass.
[tex] n_{NaCl} = n_{Na} = 1.30 moles [/tex]
[tex] m_{NaCl} = n_{NaCl}*M = 1.30 moles*58.44 g/mol = 75.9 g [/tex]
Therefore, would be formed 75.9 grams of salt.
I hope it helps you!
