Question

9. In the reaction of sodium hydroxide with chlorine gas, sodium chloride, sodium hypochlorite, and water a

reproduced. If 48.9g of chlorine gas is bubbled into a solution containing 54.2 g NaOH, how many grams
of NaClO can eventually be produced?
2 NaOH +Cl, = NaCl + NaClO + H,O

Answer 1

To calculate the amount of sodium hypochlorite (NaClO) produced in the reaction, we need to determine the limiting reactant. We convert the given masses of NaOH and Cl₂ into moles, use the balanced equation to determine the mole ratio, and find that the limiting reactant is Cl₂. Finally, we convert the moles of NaClO to grams.

Step-by-step explanation:

To calculate the amount of sodium hypochlorite (NaClO) produced in the reaction of sodium hydroxide (NaOH) with chlorine gas (Cl₂), we need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed in the reaction and determines the maximum amount of product that can be formed. To do this, we first need to convert the given masses of NaOH and Cl₂ into moles.

The molar mass of NaOH is 40 g/mol, so 54.2 g of NaOH is equal to 54.2/40 = 1.355 mol of NaOH.

The molar mass of Cl₂ is 70.90 g/mol, so 48.9 g of Cl₂ is equal to 48.9/70.90 = 0.689 mol of Cl₂.

Now, we can use the balanced equation to determine the mole ratio between NaOH and NaClO. From the equation, we can see that 2 moles of NaOH react with 1 mole of Cl₂ to produce 1 mole of NaClO.

Since the mole ratio is 2:1, the limiting reactant is Cl₂ because we have less moles of Cl₂ compared to NaOH. Therefore, all of the 0.689 mol of Cl₂ will react and form 0.689 mol of NaClO.

Finally, we can convert the moles of NaClO to grams using its molar mass. The molar mass of NaClO is 74.44 g/mol, so 0.689 mol of NaClO is equal to 0.689 × 74.44 = 51.20 g of NaClO.

Answer 2

Answer: 51.4 g of NaClO will be produced.

Step-by-step explanation:

To calculate the moles :

`\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}`

`\text{Moles of} Cl_2=(48.9g)/(71g/mol)=0.69moles`

`\text{Moles of} NaOH=(54.2g)/(40g/mol)=1.4moles`

`2NaOH+Cl_2\rightarrow NaCl+NaClO+H_2O`

According to stoichiometry :

1 moles of
`Cl_2` require = 2 moles of
`NaOH`

Thus 0.69 moles of
`Cl_2` will require=
`(2)/(1)* 0.69=1.38moles` of
`NaOH`

Thus
`Cl_2` is the limiting reagent as it limits the formation of product and
`NaOH` is the excess reagent.

As 1 mole of
`Cl_2` give = 1 mole of
`NaClO`

Thus 0.69 moles of
`Cl_2` give =
`(1)/(1)* 0.69=0.69moles` of
`NaClO`

Mass of
`NaClO=moles* {\text {Molar mass}}=0.69moles* 74.5g/mol=51.4g`

Thus 51.4 g of NaClO will be produced