Question

sodium reacts with chlorine gas to form sodium chloride. if you have 60 L of chlorine gas at STP and 30 g of sodium, how many grams of salt would be formed? i know the answer is 75 grams but i need to show my work and i don't know what to write for the equation

Answer 1

75.9 grams of salt

Step-by-step 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.

`n_(Na) = (m)/(A_(r)) = (30 g)/(22.99 g/mol) = 1.30 moles`

The number of moles of Cl₂ can be found by the Ideal gas law equation:

`PV = n_{Cl_(2)}RT`

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)

`n_{Cl_(2)} = (PV)/(RT) = (1 atm*60 L)/(0.082 L*atm/(K*mol)*273 K) = 2.68 moles`

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:

`n_(Na) = (2 moles Na)/(1 mol Cl_(2))*2.68 moles Cl_(2) = 5.36 moles`

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.

`n_(NaCl) = n_(Na) = 1.30 moles`

`m_(NaCl) = n_(NaCl)*M = 1.30 moles*58.44 g/mol = 75.9 g`

Therefore, would be formed 75.9 grams of salt.

I hope it helps you!