Question #0d5eb

Start by taking a look at the balanced chemical equation for this neutralization reaction

#color(red)(2)"NaOH"_text((aq]) + "H"_2"SO"_text(4(aq]) -> "Na"_2"SO"_text(4(aq]) + 2"H"_2"O"_text((l])#

The #color(red)(2):1# mole ratio that exists between sodium hydroxide and sodium sulfate tells you that the reaction will produce #1# mole of sodium sulfate for every #color(red)(2)# moles of sodium hydroxide that take part in the reaction.

In your case, sulfuric acid is in excess, which means that you can assume that all the mass of sodium hydroxide will take part in the reaction, since sodium hydroxide is a limiting reagent.

Now, in order to use gram to gram stoichiometry directly, you must use the molar masses of the two compounds to convert the mole ratio to a gram ratio.

Sodium hydroxide has a molar mass of #"40.0 g mol"^(-1)#, so one mole of sodium hydroxide will have a mass of #"40.0 g"#.

Sodium sulfate has a molar mass of #"142.04 g mol"^(-1)#, so one mole of sulfuric acid will have a mass of #"142.04 g"#.

The #color(red)(2):1# mole ratio will be equivalent to

#(color(red)(2) xx "40.0 g") : (1 xx "142.04 g") = "80.0 g ":" 142.04 g"#

This means that the reaction produces #"142.04 g"# of sodium sulfate for every #"80.0 g"# of sodium hydroxide consumed.

This means that #"200 g"# of sodium hydroxide will produce

#200color(red)(cancel(color(black)("g NaOH"))) * ("142.04 g Na"_2"SO"_4)/(80.0color(red)(cancel(color(black)("g NaOH")))) = "355.1 g Na"_2"SO"_4#

You must round this off to one sig fig, since that is how many sig figs you have for the mass of sodium hydroxide

#m_(Na_2SO_4) = color(green)(|bar(ul(color(white)(a/a)"400 g Na"_2"SO"_4color(white)(a/a)))|)#