Answer to Question #345622 in General Chemistry for kris

Question

350.0 mL of 0.450 mol/L sodium carbonate (Na2CO3) solution and 400.0 mL of 0.700 mol/L calcium chloride (CaCl2) solution are mixed together. Calculate the mass of calcium carbonate that would precipitate.

Answer 1

Na₂CO₃ + CaCl₂ = 2 NaCl + CaCO₃

1. Na₂CO ₃va CaCl₂ ning moddaning (mol) miqdorini topamiz. buning uchun quyidagi formuladan foydalanamiz:

n = C_M * V

in the formula,

n — amount of substance (mol);

C_M — concentration of molarity (mol/l);

V — volume of solution (litres).

a) Na₂CO₃ :

n = 0,45 mol/l * 0,35 l = 0,1575 mol

b) CaCl₂ :

n = 0,7 mol/l * 0,4 l = 0,28 mol

2. To find the product of the reaction, we use to calculate the amount of substance (mol) that is less than the reactants. because a substance with a small amount (mol) is completely consumed in the reaction, and a substance with a high content (mol) increases by a certain amount. that is, we calculate using the amount of Na₂CO₃ and find the mass of CaCO₃ :

According to above the reaction,

1mol Na₂CO₃ — 100gr (1mol) CaCO₃

0,1575 mol Na₂CO₃ — X gr CaCO₃

X = 0,1575 * 100 / 1 = 15,75 gr

ANSWER: 15,75gr CaCO₃