Question

CaCl2(aq) + Na2CO3(aq) --> CaCO3(s) + 2NaCl(aq)

20 mL of 0.5M CaCl₂ is added to 10 mL of 1.5M Na₂CO_3. Determine the number of moles of CaCl₂ and Na₂CO₃ and calculate the the number of moles of CaCO₃ that would be expected. In and experiment you produced 0.898g of CaCO₃ what is the precent yeild?

% yeild=________________

Answer 1

CaCl2(aq) + Na2CO3(aq) --> CaCO3(s) + 2NaCl(aq)

20 mL of 0.5M CaCl₂ is added to 10 mL of 1.5M Na₂CO_3.

1) Determine the number of moles of CaCl₂ and Na₂CO₃

Molarity = no. of moles/Volume of solution(L)

a) 20 mL= 0.02 L of 0.5M CaCl2 ; no. of moles = M*V(L) = 0.5*0.02 = 0.01 moles of CaCl2

b) 10 mL=0.01 L of 1.5M Na₂CO3 ; no. of moles = M*V(L) = 1.5*0.01 = 0.015 moles of Na2CO3

2) calculate the the number of moles of CaCO₃ that would be expected.

From balanced equation, 1mole of CaCl2 gives 1 moles of CaCO3. Thus 0.01 mole of CaCl2 gives 0.01 mole of CaCO3

amount of CaCO3 will be formed = moles*molar mass=0.01*101 = 1.01 gms (theoritical yield)

3) In and experiment you produced 0.898g of CaCO₃ what is the precent yeild?

%yield = (actual yield/ theoritical yield)*100

%yield = (0.898/1.01)*100 =88.91 %

% yeild= 88.91%