Question

2Na+ClX

22NaCl
How many moles of
N
a
C
l
NaClN, a, C, l will be produced from
59.0
g
59.0 g59, point, 0, start text, space, g, end text of
N
a
NaN, a, assuming
C
l
2
ClX
2

​
is available in excess?

Answer 1

2.566

Explanation:

The molar weight of Na is 22.989769 g, so 59 g represents ...

(59 g)/(22.989769 g/mol) = 2.566 mol

2.566 mol of Na will react to produce 2.566 moles of NaCl when Cl is available in excess.