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reactia 0,04 moli h clor cu Ca Co3. aflati cantitatea in grame si moli de produsii de reactie

Răspuns :

Moonlight01
2HCl + CaCO3 = CaCl2 + H2CO3 (CO2+H2O)

2 moli HCl.....1 mol CaCO3.......1 mol CaCl2........1mol CO2.....1mol H2O
0,04 moli HCl......x=0,02 moli.......y=0,02 moli........z=0,02 moli....w=0,02 moli

MCaCl2=ACa+2ACl=40+2*35,5=111g/mol

MCO2=AC+2AO=12+2*16=44g/mol

MH2O=2AH+AO=2*1+16=18g/mol

n=m/M=>m=n*M=0,02*111=2,22g CaCl2

n=m/M=>m=n*M=0,02*44=0,88g CO2

n=m/M>m=n*M=0,02*18=0,36g H2O

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