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?

1 Answer
anor277
Aug 21, 2017

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

Explanation:

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#.

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