2NaCl(aq) + 2H20(l) 2Na0H(aq) + Cl2(g) Find the number of moles of sodium chloride needed to produce 8.0 L of chlorine gas at STP.

Answer 1

At STP conditions 1 mol of gas occupie 22.4 L. Then 8 L of Cl2 represent 8 divided by 22.4, 0.3571 moles.

From the balanced equation we need 2 moles of NaCl to obtain 1 Mol of Cl2, Then we need 2 times 0.3571 moles of NaCl. this is 0.7142 moles of NaCl

Answer 2

This question is asking us to find the number of moles of NaCl that we would need to produce 8.0L of Cl₂ gas at STP, and it gives us the balanced chemical equation that describes the reaction.

2NaCl + 2H2O ----> 2NaOH + Cl2

From looking at the information in the question, we can see that we are given a value for volume which we can call V. We also were told that the container that the gas was in was at STP, or Standard Temperature and Pressure which always means that the gas is at 0°C, or 273.15K and 1 atmosphere (atm) of pressure.

Knowing this, we can make a list of everything we have and see where to go from that

P = 1atm

V = 8L

moles (n) = ?

T = 273K

With all of these values and and the fact that we are dealing with Cl2 gas, we can use the ideal gas law, PV=nRT to solve for the amount of Cl2 gas that was generated. We should also note that since we are using the units L, atm, mol, we will use the gas constant R = 0.08206

PV=nRT

PV = n

RT

1atm X 8L = n

0.08206((L*atm)/(mol*K)) X 273K

n = 0.3571 mol Cl2 gas

From this, we can use the chemical equation to find the number of moles of NaCl needed to create this gas. In the equation

2NaCl + 2H2O ----> 2NaOH + Cl2

we can see that for every one mole of Cl2, there are 2 moles of NaCl. There are twice as many moles. Therefore, all we have to do is multiply the number of moles by two to get the answer!

0.3571 mol Cl2 X ((2 mol NaCl)/(1 mol Cl2)) = 0.7142 mol NaCl

Finally, to correct for significant figures, we are left with 0.71 moles of NaCl needed to create 8.0L of Cl2 gas