Answer
$$ 3.1 \space g \space NH_4SNC $$
Work Step by Step
a.
Balance the amount of $S$ in each side:
$$Ba(OH)_2 \space 8 H_2O(s) + 2 \space NH_4SCN(s) \longrightarrow Ba(SCN)_2(s) + H_2O(l) + NH_3(g)$$
Balance the amount of $N$ in each side:
$$Ba(OH)_2 \space 8 H_2O(s) + 2 \space NH_4SCN(s) \longrightarrow Ba(SCN)_2(s) + H_2O(l) + 2 \space NH_3(g)$$
Balance the amount of $O$:
$$Ba(OH)_2 \space 8 H_2O(s) + 2 \space NH_4SCN(s) \longrightarrow Ba(SCN)_2(s) + 10 \space H_2O(l) + 2 \space NH_3(g)$$
b.
$ Ba(OH)_2 \space 8H_2O $ : ( 137.3 $\times$ 1 )+ ( 1.008 $\times$ 18 )+ ( 16.00 $\times$ 10 )= 315.40 g/mol
$$ \frac{1 \space mole \space Ba(OH)_2 \space 8H_2O }{ 315.40 \space g \space Ba(OH)_2 \space 8H_2O } \space and \space \frac{ 315.40 \space g \space Ba(OH)_2 \space 8H_2O }{1 \space mole \space Ba(OH)_2 \space 8H_2O }$$
$ NH_4SNC $ : ( 1.008 $\times$ 4 )+ ( 12.01 $\times$ 1 )+ ( 14.01 $\times$ 2 )+ ( 32.07 $\times$ 1 )= 76.13 g/mol
$$ \frac{1 \space mole \space NH_4SNC }{ 76.13 \space g \space NH_4SNC } \space and \space \frac{ 76.13 \space g \space NH_4SNC }{1 \space mole \space NH_4SNC }$$
$$ 6.5 \space g \space Ba(OH)_2 \space 8H_2O \times \frac{1 \space mole \space Ba(OH)_2 \space 8H_2O }{ 315.40 \space g \space Ba(OH)_2 \space 8H_2O } \times \frac{ 2 \space moles \space NH_4SNC }{ 1 \space mole \space Ba(OH)_2 \space 8H_2O } \times \frac{ 76.13 \space g \space NH_4SNC }{1 \space mole \space NH_4SNC } = 3.1 \space g \space NH_4SNC $$
