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What is the mass of potassium iodide (166.00 g/mol) that yields 0.500 g of lead(ii) iodide (461.0 g/mol) precipitate? (you must balance the equation first) __pb(no3)2(aq) + __ki(s) → __pbi2(s) + __kno3(aq) 0.0900 g 2.78 g 0.694 g 0.180 g 0.360 g?

Respuesta :

kenmyna

 the  mass  of  potassium iodide  is  0.360  g


calculation

 step 1:  write the balanced molecular  equation

Pb(NO3)2  (aq)  + 2 Ki (s)→ Pbi2(s)  + 2KNO3

step 2 ; calculate the  moles of Pbi2

moles =  mass /molar mass

= 0.500 g / 461 .0  g/mol  =0.0011 moles


step  3:  use the mole ratio to  calculate  the  moles of Ki

Ki: Pbi2  is  2:1  therefore  the  moles of  ki  = 0.0011  x 2/1 =  0.0022  moles


step 4 :  find the mass of Ki

mass=  moles  x molar  mass

=0.0022  moles  x 166g/mol  =0.365  g  which is  approximate  0.360 g

0.360 grams is the mass of potassium iodide (KI).

How we calculate mass from moles?

Mass of any substance will be calculated from moles as:

n = W/M, where

W = required mass

M = molar mass

Given chemical reaction is:

Pb(NO₃)₂(aq)  + 2KI(s) → Pbi₂(s) + 2KNO₃

From the stoichiometry of the solution, it is clear that

2 moles of KI = produce 1 mole of Pbi₂

Moles of Pbi₂ = 0.500 g / 461 .0  g/mol = 0.0011 moles

0.0011 moles of Pbi₂ = produced by 2×0.0011 = 0.0022 moles of KI

Now we calculate the mass of KI using the above formula of moles:

mass = moles x molar mass

mass = 0.0022moles x 166g/mol = 0.365g = 0.360g

Hence, option (5) is correct.

To know more about moles, visit the below link:
https://brainly.com/question/15263827

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