Tính % các nguyên tố hóa học trong mỗi hợp chất sau a) Ca(NO3)2 b) Zn(OH)2 c) KNO3 d) Na2CO3 ( TÍNH THEO CÁCH CỦA CHƯƠNG TRÌNH MỚI Ạ)

`a)`

`M_{Ca(NO_3)_2}=40+(14+16.3).2=164(a m u)`

`\%m_{Ca}={40}/{164}.100\%\approx 24,39\%`

`\%m_N={14.2}/{164}.100\%\approx 17,07\%`

`\%m_O=100-17,07-24,39=58,54\%`

`b)`

`M_{Zn(OH)_2}=65+(16+1).2=99(a m u)`

`\%m_{Zn}={65}/{99}.100\%\approx 65,66\%`

`\%m_O={16.2}/{164}.100\%\approx 32,32\%`

`\%m_H=100-65,66-32,32=2,02\%`

`c)`

`M_{KNO_3}=39+14+16.3=101(a m u)`

`\%m_{K}={39}/{101}.100\%\approx 38,61\%`

`\%m_N={14}/{101}.100\%\approx 13,86\%`

`\%m_O=100-38,61-13,86=47,53\%`

`d)`

`M_{Na_2CO_3}=23.2+12+16.3=106(a m u)`

`\%m_{Na}={23.2}/{106}.100\%\approx 43,4\%`

`\%m_C={12}/{106}.100\%\approx 11,32\%`

`\%m_O=100-43,4-11,32=45,28\%`

$\rm a)\\M_{Ca(NO_3)_2}=40+(14+16.3).2=164(a m u)\\\xrightarrow[]{}\\\%m_{Ca}=({40}/{164}).100\%\approx 24,39\%\\\%m_N=({14.2}/{164}).100\%\approx 17,07\%\\\%m_O=100-17,07-24,39=58,54\%\\b)\\M_{Zn(OH)_2}=65+(16+1).2=99(a m u)\\\xrightarrow[]{}\\\%m_{Zn}=({65}/{99}).100\%\approx 65,66\%\\\%m_O=({16.2}/{164}).100\%\approx 32,32\%\\\%m_H=100-65,66-32,32=2,02\%\\c)\\M_{KNO_3}=39+14+16.3=101(a m u)\\\xrightarrow[]{}\\\%m_{K}=({39}/{101}).100\%\approx 38,61\%\\\%m_N=({14}/{101}).100\%\approx 13,86\%\\\%m_O=100-38,61-13,86=47,53\%\\d)\\M_{Na_2CO_3}=23.2+12+16.3=106(a m u)\\\xrightarrow[]{}\\\%m_{Na}=({23.2}/{106}).100\%\approx 43,4\%\\\%m_C=({12}/{106}).100\%\approx 11,32\%\\\%m_O=100-43,4-11,32=45,28\%$