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

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.

$n_{Na} = \frac{m}{A_{r}} = \frac{30 g}{22.99 g/mol} = 1.30 moles$

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)

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

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:

$n_{Na} = \frac{2 moles Na}{1 mol Cl_{2}}*2.68 moles Cl_{2} = 5.36 moles$

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!