Answer to Question #345834 in General Chemistry for Jonalyn

Question

1.)calculate the volume, in mL of 1.72M HCl solution that will react with 2.67 mol of CaCO3? (Ans. 3.10x103mL

___HCl+___CaCO3____CaCl2+___H2O+__CO2

2.)If 0.650% solution of NaCO3 will be used to precipitate Ca+2 from solution, determine the amount of solution, in grams, needed to precipitate 5.00x102mL of 0.01120M Ca+2?(Atomic mass :Na=22.99g/mol:C=12.01g/mol:O=16.00g/mol:Ca=40.08g/mol:Ans.91.3g)

____Na2CO3(aq)+___Ca+2(aq)____CaCO3(s)+___Na+(aq)

Answer 1

1).

CaCO3 + 2 HCl = CaCl₂ + CO₂ + H2O

According to the above reaction,

1 mol CaCO₃ —> 2 mol HCl

2,76mol CaCO₃ —> X

X = 2,76 * 2 / 1 = 5,52 mol HCl

V = n / C_M = 5,52mol / 1,72mol/l = 3,21 litr or 3,21 * 10³ ml

V — volume of solution (litr);

n — amount of substance (mol);

C_M — concentration of molarity.

ANSWER: 3,21 * 10³ ml.

2).

a)

V — volume = 5*10²ml = 0,5 litr;

C_M — concentration of molarity =

= 0,0112 mol/l;

n — amount of substance (mol) = ?

n (Ca²⁺) = V * C_M = 0,5 * 0,0112 =0,0056 mol

b)

M_r (Na₂CO₃) = 23*2+12+16*3 = 106 g/mol

Na₂CO₃ + Ca²⁺ => CaCO₃ + 2Na⁺

According to the above reaction,

106 g Na₂CO₃ — > 1mol Ca²⁺

X — > 0,0056mol Ca²⁺

X = 106 * 0,0056 / 1 = 0,5936 g Na₂CO₃

c)

0,5936 g — 0,65%

X — 100%

X = 0,5936 * 100 / 0,65 = 91,3 g

ANSWER: 91,3 g