Question

How many grams of Br are required to react completely with 29.5 g of NO? 2NO+Br2- 2NOBr

Answer 1

Answer: 78.54 grams

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

$\text{Number of molesof} NO=\frac{29.5g}{30.01g/mol}=0.98moles$

$2NO+Br_2\rightarrow 2NOBr$

According to stoichiometry :

2 moles of $NO$ react with 1 mole of $Br_2$

0.98 moles of $NO$ react with =$\frac{1}{2}\times 0.98=0.49$ moles of $Br_2$

Mass of $Br_2=moles\times {\text{Molar mass}=0.49\times 159.8=78.54g$

Thus 78.54 g of $Br_2$ are required to react completely with 29.5 g of NO.