#### Answer

(a) $ 8.00mol(O_2)$
(b) $6.40mol(NO)$
(c) $9.60mol(H_2O)$

#### Work Step by Step

1. Balance the equation:
- Begin with the elements that appear only once in each side:
Balance the $H$:
The subscript for the hydrogen on the reactants is "3" from: $NH_3$. So, we should put a "$\frac{3}{2}$" as the coefficient of $H_2O$.
* We divided by 2, because each water molecule has 2 hydrogens.
$NH_3 + O_2 -- \gt NO + \frac{3}{2}H_2O$
Multiply all coefficients by 2:
$2NH_3 + 2O_2 -- \gt 2NO + 3H_2O$
Balance the $O$:
The products side has a total of 5 oxygens. Therefore we should put a "$\frac{5}{2}$" as the coefficient of $O_2$:
$2NH_3 + \frac{5}{2}O_2 -- \gt 2NO + 3H_2O$
Multiply all coefficients by 2:
$4NH_3 + 5O_2 -- \gt 4NO + 6H_2O$
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Use the balanced coefficients as conversion factors:
(a) $6.40mol(NH_3) \times \frac{5(O_2)}{4(NH_3)} = 8.00mol(O_2)$
(b) $6.40mol(NH_3) \times \frac{4(NO)}{4(NH_3)} = 6.40mol(NO)$
(c) $6.40mol(NH_3) \times \frac{6(H_2O)}{4(NH_3)} = 9.60mol(H_2O)$