Question #19122

1 Answer
Stefan V.
Jan 29, 2015

#"5.25 g"# of #"NaF"# will be produced (or #"5 g"# if you round to 1 sig fig - the number of sig figs 0.5 has).

So, you have your balanced chemical equation, which I won't write again here. Notice that you have an #"8:2"#, or better said, a #"4:1"# mole ratio between #"HF"# and #"NaF"#.

This means that #"4 moles"# of #"HF"# will produce #"1 mole"# of #"NaF"#. Since you were given the number of #"HF"# moles, and told that #Na_2SiO_3# was not a limiting reagent, you can determine the number of #"NaF"# moles to be

#"0.5 moles HF" * ("1 mole NaF")/("4 moles HF") = "0.125 moles"#

#"NaF"# has a molar mass of #"42.0 g/mol"#, which means that the mass produced will be

#"0.125 moles" * ("42.0 g")/("1 mole") = "5.25 g NaF"#

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