Respuesta :
Answer:
1) [tex]K_{2}CO_{3}(aq)+CaCl_{2}(aq)\rightarrow CaCO_{3}(s)+2KCl(aq)[/tex]
2) K2CO3 is the limiting reagent
3) Mass of K2CO3 need = 4.67 g
Explanation:
1) Potassium carbonate reacts with calcium chloride to form calcium carbonate and potassium chloride. The balanced equation is:
[tex]K_{2}CO_{3}(aq)+CaCl_{2}(aq)\rightarrow CaCO_{3}(s)+2KCl(aq)[/tex]
Based on the reaction stoichiometry:
1 mole of Potassium carbonate reacts with 1 mole of calcium chloride to form 1 mole of calcium carbonate and 2 moles potassium chloride
2) Mass of CaCl2 = 15 g
Molar mass of CaCl2 = 110.98 g/mol
[tex]Moles\ CaCl2 = \frac{Mass\ CaCl2}{Molar\ Mass} =\frac{15g}{110.98 g/mol}=0.135[/tex]
Mass of K2CO3[tex]Mass\ of\ K2CO3 = moles*molar\ mass=0.0338moles*138.21g/mol=4.67 g[/tex] = 15 g
Molar mass of K2CO3 = 138.21 g/mol
[tex]Moles\ K2CO3 = \frac{Mass\ K2CO3}{Molar\ Mass} =\frac{15g}{138.21g/mol}=0.109[/tex]
Since the moles of K2CO3 < moles of CaCl2, potassium carbonate is the limiting reagent
3) 15% CaCl2 implies: 15 g CaCl2 in 100 ml solution
Therefore in the given 25 ml solution, the mass of CaCl2 would be:
[tex]\frac{25ml*15g}{100 ml} =3.75 g[/tex]
[tex]Moles\ CaCl2 = \frac{Mass\ CaCl2}{Molar\ Mass} =\frac{3.75g}{110.98 g/mol}=0.0338[/tex]
From the reaction stoichiometry the molar ratio of K2CO3:CaCl2 = 1:1
Therefore, moles of K2CO3 that would be needed to fully precipitate 0.0338 moles of CaCl2 present in 25 ml of 15% solution would be = 0.0338 moles
The corresponding mass of K2CO3 would be:
[tex]Mass\ K2CO3 = moles*molar\ mass = 0.0338moles*138.21g/mol=4.67g[/tex]
The mass of the calcium chloride is 3.75 g.
A chemical reaction equation must consist of the reactant side and the product side.
Now the reaction as specified is;
K2CO3(aq) + CaCl2(aq) ----> CaCO3(s) + 2KCl(aq)
Number of moles of K2CO3 = 15 g/138 g/mol = 0.11 moles
Number of moles of CaCl2 = 15 g/111 g/mol = 0.14 moles
Since the reaction is 1:1, it follows that K2CO3 is the limiting reactant.
A 15% solution means 15g of CaCl2 in a 100 mL solution hence;
If 15 g is contained in 100 mL of solution
x g is contained in 25 mL of solution
X = 3.75 g
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