Question

Se trateaza 71 grame de Na2SO4 cu BaCl2. Ce produsi rezulta si in ce cantitati. rezultatele sa se exprime in grame si moli.

Answer 1

MNa2SO4 = 2ANa+AS+4AO=2*23+32+4*16=142g/mol

MBaCl2=ABa+2ACl=137+2*35,5=208g/mol

MNaCl=ANa+ACl=23+35,5=58,5g/mol

MBaSO4=ABa+AS+4AO=137+32+4*16=233g/mol

n=m/M=71/142=0,5 moli Na2SO4

Na2SO4 + BaCl2 ---> 2NaCl + BaSO4

1 mol Na2SO4.................2 moli NaCl........1mol BaSO4
0,5 moli Na2SO4.............x=1 mol...............y=0,5 moli

1 mol Na2SO4.....................58,5g NaCl........233g BaSO4
0,5 moli Na2SO4..................x=29,25g NaCl......y=116,5g BaSO4