Question

$2NO+B{r}_2⟶2NOBr$
For every $5.0$ moles of bromine that reacts, the sum of moles of $NO$ reacted and moles of $NOBr$ is :

Answer 1

$1$ mole of $B{r}_2$ reacts with $2$ moles of $NO$ to produce $2$ moles of $NOBr$.
Hence, $5.0$ moles of $B{r}_2$ will react with $10.0$ moles of $NO$ to produce $10.0$ moles of $NOBr$.
Hence, the sum of the number of moles of $NO$ reacted and the number of moles of $NOBr$ formed is $10+10=20$.