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If 70.0g of manganese (IV) fluoride reacts with 70.0g ammonium sulfide, which reactant is
the limiting reactant and how many grams of the excess reactant is left over?
MnF_4(aq)+2(NH_4)_2S(aq)-> MnS_2(s)+4NH_4F(aq)

Sagot :

Oseni

The limiting reactant would be the ammonium sulfide and the amount of the excess manganese (IV) fluoride would be 1.96 grams

Limiting reactants

From the equation of the reaction:

[tex]MnF_4(aq)+2(NH_4)_2S(aq)- > MnS_2(s)+4NH_4F(aq)[/tex]

The mole ratio of [tex]MnF_4[/tex] to [tex](NH_4)_2S[/tex] is 1:2

Mole of 70 g [tex]MnF_4[/tex] = 70/130.93

                                 = 0.53 moles

Mole of 70 g  [tex](NH_4)_2S[/tex] = 70/68.154

                                       = 1.03 moles

Equivalent mole of   [tex](NH_4)_2S[/tex] = 0.53 x 2

                                                = 1.06 moles

Thus,   [tex](NH_4)_2S[/tex] seems to be limiting while [tex]MnF_4[/tex] is in excess.

Excess [tex]MnF_4[/tex] = 0.53 - 1.03/2

                          = 0.015 mole

Mass of 0.015 mole [tex]MnF_4[/tex] = 0.015 x 130.93

                                             = 1.96 grams

More on limiting reactants can be found here: https://brainly.com/question/14225536

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