Đáp án:
Giải thích các bước giải:
a/ Nhóm (NO3) có hoá trị (I)
KNO3 có PTK: 39+14+3.16 = 101
→ % K = $\frac{39}{101}$ .100 = 38,61%
→ % N = $\frac{14}{101}$ .100 = 13,86%
→ % O = $\frac{3.16}{101}$ .100 = 47,52%
Mg(NO3)2 có PTK: 24+2.(14+3.16) = 148
→ % Mg = $\frac{24}{148}$ .100 = 16,22%
→ % N = $\frac{2.14}{148}$ .100 = 18,92%
→ % O = $\frac{2.3.16}{148}$ .100 = 64,86%
Fe(NO3)3 có PTK: 56+3.(14+3.16) = 242
→ % Fe = $\frac{56}{242}$ .100 = 23,14%
→ % N = $\frac{3.14}{242}$ .100 = 17,36%
→ % O = $\frac{3.3.16}{242}$ .100 = 59,5%
b/ Nhóm (PO4) có hoá trị (III)
K3PO4 có PTK: 3.39+31+4.16 = 212
→ % K = $\frac{3.39}{212}$ .100 = 55,19%
→ % P = $\frac{31}{212}$ .100 = 14,62%
→ % O = $\frac{4.16}{212}$ .100 = 30,19%
Mg2(PO4)3 có PTK: 2.24+3.(31+4.16) = 333
→ % Mg = $\frac{2.24}{333}$ .100 = 14,41%
→ % P = $\frac{3.31}{333}$ .100 = 27,93%
→ % O = $\frac{3.4.16}{333}$ .100 = 57,66%
FePO4 có PTK: 56+31+4.16 = 151
→ % Fe = $\frac{56}{151}$ .100 = 37,09%
→ % P = $\frac{31}{151}$ .100 = 20,53%
→ % O = $\frac{4.16}{151}$ .100 = 42,38%
c/ Nhóm (OH) có hoá trị (I)
KOH có PTK: 39+16+1 = 56
→ % K = $\frac{39}{56}$ .100 = 69,64%
→ % O = $\frac{16}{56}$ .100 = 28,57%
→ % H = $\frac{1}{56}$ .100 = 1,79%
Mg(OH)2 có PTK: 24+2.(16+1) = 58
→ % Mg = $\frac{24}{58}$ .100 = 41,38%
→ % O = $\frac{2.16}{58}$ .100 = 55,17%
→ % H = $\frac{2.1}{58}$ .100 = 3,45%
Fe(OH)3 có PTK: 56+3.(16+1) = 107
→ % Fe = $\frac{56}{107}$ .100 = 52,34%
→ % O = $\frac{3.16}{107}$ .100 = 44,86%
→ % H = $\frac{3.1}{107}$ .100 = 2,8%
