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

Step-by-step explanation:

According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number
`6.023* 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=(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 =
`(1)/(2)* 0.98=0.49` moles of
`Br_2`

Mass of
`Br_2=moles* {\text{Molar mass}=0.49* 159.8=78.54g`

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