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Tin (II) fluoride , SnF2, is found in some toothpastes. Tin (III) fluoride is made by reacting solid tin with hydrogen fluoride according to the following BALANCED equation. Sn(s) + 2 HF (g) -> SnF2(s) + H2(g) How many moles of tin are needed to react with 8.4 moles of hydrogen fluoride ?

Respuesta :

transitionstate
The Balance Chemical Equation is as;

                                 Sn  +  2 HF   →    SnF₂  +  H₂

According to Equation,

                        2 Moles of HF requires  =  1 Mole of Sn
So,
                 8.4 Moles of HF will require  =  X Moles of Sn

Solving for X,
                            X  =  (8.4 mol × 1 mol) ÷ 2 mol

                            X  =  4.2 Moles of Sn

Result:
           4.2 Moles of Tin are needed to react with 8.4 moles of Hydrogen Fluoride
shristiparmar1221

The number of moles of tin needed to react with 8.4 moles of hydrogen fluoride are 4.2 Moles. It can be founded with the help of limiting reagent concept.

What is Limiting reagent ?

The limiting reactant (or limiting reagent) is the reactant that gets consumed first in a chemical reaction and therefore limits how much product can be formed.

Given Balance Chemical Equation ;

                                Sn (s)  +  2 HF (g)   →    SnF₂ (s)  +  H₂ (g)

According to Equation,

                      2 Moles of HF requires  =  1 Mole of Sn

Therefore,

                8.4 Moles of HF will require  =  X Moles of Sn

Solving for X,

                           X  =  (8.4 mol × 1 mol) ÷ 2 mol

                           X  =  4.2 Moles of Sn

Hence, The number of moles of tin needed to react with 8.4 moles of hydrogen fluoride are 4.2 Moles.

Learn more about Limiting reagent here ;

https://brainly.com/question/20070272

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