Question #0cbef
1 Answer
Explanation:
The key to this problem is the balanced chemical equation of the reaction.
Chlorine gas,
#"Cl"_text(2(g]) + 2"NaOH"_text((aq]) -> "NaClO"_text((aq]) + "NaCl"_text(aq]) + "H"_2"O"_text((l])#
Notice that you ahve a
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_"NaClO" = "40 g"#