Answer:
0.143 g of KCl.
Explanation:
Equation of the reaction:
AgNO3(aq) + KCl(aq) --> AgCl(s) + KNO3(aq)
Molar concentration = mass/volume
= 0.16 * 0.012
= 0.00192 mol AgNO3.
By stoichiometry, 1 mole of AgNO3 reacts with 1 mole of KCl to form a precipitate.
Number of moles of KCl = 0.00192 mol.
Molar mass of KCl = 39 + 35.5
= 74.5 g/mol
Mass = molar mass * number of moles
= 74.5 * 0.00192
= 0.143 g of KCl.
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Answer:
14.33 g
Explanation:
Solve this problem based on the stoichiometry of the reaction.
To do that we need the molecular weight of the masses involved and then calculate the number of moles, find the limiting reagent and finally calculate the mass of AgCl.
2 AgNO₃ + CaCl₂ ⇒ Ca(NO₃)₂ + 2 AgCl
mass, g 6.97 6.39 ?
MW ,g/mol 169.87 110.98 143.32
mol =m/MW 0.10 0.06 0.10
From the table above AgNO₃ is the limiting reagent and we will produce 0.10 mol AgCl which is a mass :
0.10 mol x 143.32 g/mol = 14.33 g
Because the water seeped through the bag (? I don’t know the context correct me if I’m wrong)
