Question

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Answer 1

Răspuns:

Explicație:

ms= 200g sol. NaOH

c= 20%

CO2

msare

-se afla  md  sol. de NaOH

md= c.ms : 100

md= 20  . 200  : 100=40g NaOH

40g                       xg

2NaOH + CO2= Na2CO3 + H2O

2 .40g                 106g

x= 40. 106 : 2 .40=53g Na2CO3

n = 53 g : 106g/moli=0,5 moli sare

MNaOH= 23+16 +1=40------> 40g/moli

MNa2CO3=2.23 + 12 + 3.16=106------> 106g/moli

Answer 2

ms = 200g, c% = 20%

c% = mdx100/ms

=> md = msxc%/100

= 200x20/100 = 40 g NaOH

M.NaOH = 23+16+1 = 40 g/mol

M.Na2CO3 = 2x23+12+3x16 = 106 g/mol

40g                           m g

2NaOH + CO2 --> Na2CO3 + H2O

2x40                       106

=> m = 40x106/2x40 = 53 g Na2CO3

nr.moli = m/M = 53/106 = 0,5 moli Na2CO3