Rate law equation - orders? Consider the following reaction:
  
  $$\ce{2 H2 (g) + 2 NO (g) -&gt; N2 (g) + 2 H2O (g)}$$
  
  The elementary steps of this reaction are:
  
  $$
\begin{align}
&amp;1^\text{st}~\text{step:} &amp;\ce{H2 (g) + 2 NO (g) &amp;-&gt; N2O (g) + H2O (g)} &amp;&amp;\text{(slow)} \\
&amp;2^\text{nd}~\text{step:} &amp;\ce{N2O (g) + H2 (g) &amp;-&gt; N2 (g) + H2O (g)} &amp;&amp;\text{(fast)} \\
\end{align}
$$
  
  The reaction is exothermic overall.
  
  (a) What is the rate law for this reaction? Justify your answer.
  
  $$\text{rate} = k[\ce{NO}]^2[\ce{H2}]$$
  
  The 1st step of the reaction is the rate limiting step.


How did they know the order of the reaction with respect to $\ce{NO}$ was $2$, and $1$ for $\ce{H2}$? With other questions the orders are something they state in the question and are unrelated to the coefficients.
You are correct, the reaction order cannot be deduced from the brutto-equation and is usually determined experimentally.

However, here you are presented with the elementary reactions and their rates. Obviously, the slowest step is the bottleneck and controls the overall kinetics, hence the total rate matches the rate of the slowest step. And as this is an elementary reaction, you just apply the law of mass action and use the stoichiometric coefficients for the rate equation.