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

51.4 g of NaClO will be produced.

Explanation:

To calculate the moles :

According to stoichiometry :

1 moles of require = 2 moles of

Thus 0.69 moles of will require=  of

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

As 1 mole of give = 1 mole of

Thus 0.69 moles of give =  of

Mass of

Thus 51.4 g of NaClO will be produced

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