Question

In the following reaction, how many grams of NaBr will react with 311 grams of Pb(NO3)2?

pb(no3)2(aq)+2NaBr(g) -> PbBr2(s) + 2NaNO3(aq)
The molar mass of NaBr is 102.9 grams and that of Pb(NO3)2 is 331.21 grams.

Answer 1

193.24g

Step-by-step explanation:

Step 1:

The balanced equation for the reaction. This is illustrated below:

Pb(NO3)2(aq) + 2NaBr(g) -> PbBr2(s) + 2NaNO3(aq)

Step 2:

Determination of the masses of Pb(NO3)2 and NaBr that reacted from the balanced equation. This is illustrated below:

Molar Mass of Pb(NO3)2 = 331.21g/mol

Molar Mass of NaBr = 102.9g/mol

Mass of NaBr from the balanced equation = 2 x 102.9 = 205.8g.

From the balanced equation,

Mass of Pb(NO3)2 that reacted = 331.21g

Mass of NaBr that reacted = 205.8g

Step 3:

Determination of the mass of NaBr that reacted with 311g of Pb(NO3)2. This is illustrated below:

From the balanced equation above,

331.21g of Pb(NO3)2 reacted with 205.8g of NaBr.

Therefore, 311g of Pb(NO3)2 will react with = (311x205.8)/331.21 = 193.24g of NaBr.

From the calculations made above, 193.24g of NaBr will react with 311g of Pb(NO3)2

Answer 2

Answer: 193.4 grams

Explanation: According to avogadro's law, 1 mole of every substance weighs equal to the molecular mass and contains avogadro's number
`6.023* 10^(23)` of particles.

`Pb(NO_3)_2(aq)+2NaBr(g)\rightarrow PbBr_2(s)+2NaNO_3(aq)`

To calculate the moles, we use the equation:

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

`{\text{Number of moles} of Pb(NO_3)_2}=(311g)/(331.21gmol)=0.94moles`

According to stoichiometry

1 mole of
`Pb(NO_3)_2` reacts with 2 moles of
`NaBr`

0.94 moles of
`Pb(NO_3)_2` will react with=
`(2)/(1)* 0.94=1.88moles` of
`NaBr`

Mass of
`NaBr=moles* {\text {molar mass}}=1.88* 102.9=193.4g`