Question

Given a `3.00*g` mass of sodium metal, what mass of chlorine gas is equivalent? What mass of sodium chloride could be made given this quantity?

Answer 1

We assume a stoichiometric excess of chlorine gas......and we work out the mass of sodium chloride produced given 100% yield.

Explanation:

....and we must write a stoichiometric equation....

`Na(s) + 1/2Cl_2(g) rarr NaCl(s)`

As written, this is a redox reaction, and sodium metal is oxidized by half an equiv of (binculear) chlorine gas....

`"Moles of natrium"=(3.00*g)/(22.99*g*mol^-1)=0.131*mol`

And thus we should get `0.131*mol` of salt.....i.e.

`0.131*molxx58.44*g*mol^-1=7.63*g`.