If you start with 5.5 grams of sodium fluoride, how many grams of magnesium fluoride will be produced in the reaction #Mg + 2NaF -> MgF_2 + 2Na#?

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Answer 1 · 2016-04-28T18:01:31

There will be 4.1 g of #"MgF"_2# produced.

Mass of sodium fluoride = 5.5 g

#"moles" = "mass"/"molar mass"#

#"moles" = "5.5 g"/"(22.99 + 18.998) g/mole"#

#"moles" = 5.5/41.998 "moles"#

#"moles" = "0.13096 moles of NaF"#

For every 2 moles of #"NaF"#, there is 1 mole of #"MgF"_2#.

Using this knowledge, we can determine the amount of moles of #"MgF"_2#.

#"0.13096 moles of NaF"/2 = "moles of MgF"_2#

#"moles of MgF"_2 = "0.06548 moles"#

We are now going to convert this into grams.

#"moles" = "mass"/"molar mass"#

#"0.06548 moles" = "mass"/"(24.305+2(18.998)) g/mole"#

#0.06548 = "mass"/"62.301 g"#

#"mass of MgF"_2 = "4.1 grams"#

Therefore, there will be 4.1 grams of #"MgF"_2# produced.