What mass of salt will be obtained when a #18.0*g# mass of sodium reacts with a #23.0*g# mass of chlorine gas?

Question

Wyatt Adkins · Accepted Answer

Approx. #38*g#............

We interrogate the chemical reaction..... #Na(s) + 1/2Cl_2(g) rarr NaCl(s)# And clearly, the stoichiometry is #1:1/2#, where we deal with #Cl_2# (you simply have to know that the elemental gases, save the Noble gases, are BINUCLEAR, i.e. #X_2#) So we take the molar quantities by dividing each mass thru by its atomic mass, which we get from the Periodic Table: #"Moles of sodium"=(18.0*g)/(22.99*g*mol^-1)=0.783*mol.# #"Moles of chlorine"=(23.0*g)/(70.9*g*mol^-1)=0.320*mol.# Given these molar quantities, it is CLEAR that chlorine gas is in deficiency, and ONLY #0.640*mol# of sodium will react. Is it clear? Why so? And thus, given complete reaction, we should get #0.640*mol# sodium chloride, which represents a mass of....... #0.640*molxx58.44*g*mol^-1~=38*g# with respect to #NaCl#.

Cooper Adams · Answer

undefined To find the mass of salt obtained when 18.0 g of sodium reacts with 23.0 g of chlorine gas, first determine the limiting reactant. Then, use the stoichiometry of the balanced chemical equation to find the mass of salt produced.

The balanced chemical equation for the reaction between sodium and chlorine gas to form salt (sodium chloride) is:

$2 \text{Na} + \text{Cl}_2 \rightarrow 2 \text{NaCl}$

Using the molar masses of sodium (22.99 g/mol) and chlorine (35.45 g/mol), calculate the number of moles of each reactant. The reactant that produces the lesser amount of product is the limiting reactant.

Moles of sodium: $18.0 \, \text{g} / 22.99 \, \text{g/mol} = 0.783 \, \text{mol}$

Moles of chlorine: $23.0 \, \text{g} / 35.45 \, \text{g/mol} = 0.649 \, \text{mol}$

Since chlorine is the limiting reactant, it determines the amount of salt produced.

From the balanced equation, 1 mole of chlorine reacts to produce 2 moles of sodium chloride. Therefore, 0.649 moles of chlorine will produce $0.649 \, \text{mol} \times 2 = 1.298 \, \text{mol}$ of sodium chloride.

Finally, calculate the mass of sodium chloride produced using its molar mass (58.44 g/mol):

$1.298 \, \text{mol} \times 58.44 \, \text{g/mol} = 75.87 \, \text{g}$

So, 75.87 g of salt will be obtained.