Question 0cbef

Oct 28, 2015

$\text{40 g}$

Explanation:

The key to this problem is the balanced chemical equation of the reaction.

Chlorine gas, ${\text{Cl}}_{2}$, is dissolved in aqueous solution and reacted with sodium hydroxide to give the hypochlorite anion, ${\text{ClO}}^{-}$, as part of sodium hypochlorite, $\text{NaClO}$, sodium chloride, $\text{NaCl}$, and water.

${\text{Cl"_text(2(g]) + 2"NaOH"_text((aq]) -> "NaClO"_text((aq]) + "NaCl"_text(aq]) + "H"_2"O}}_{\textrm{\left(l\right]}}$

Notice that you ahve a $1 : 1$ mole ratio between chlorine gas and sodium hypochlorite. This means that the reaction will produce one mole of the latter for every mole of the former that reacts.

Since you know that sodium hydroxide is in excess, you know that all the moles of chlorine gas will react. This means that the reaction will produce

0.6color(red)(cancel(color(black)("moles Cl"_2))) * "1 mole NaClO"/(1color(red)(cancel(color(black)("mole Cl"_2)))) = "0.6 moles NaClO"

Now use sodium hypochlorite's molar mass to determine how many grams would contain this many moles

0.6color(red)(cancel(color(black)("moles"))) * "74.44 g"/(1 color(red)(cancel(color(black)("mole")))) = "44.7 g"#

You need to round this off to one sig fig, the number of sig figs you gave for the number of moles of chlorine gas, so the answer will be

${m}_{\text{NaClO" = "40 g}}$