Đáp án:
Em tham khảo nhé
Giải thích các bước giải:
a. CTHH: $C_2H_6O_4Mg$
PTK = 142; $\%C=\dfrac{4.12}{142}.100\%=33,8\%$
$\%H=\dfrac{6}{142}.100\%=4,2\%$
$\%O=\dfrac{4.16}{142}.100\%=45,1\%$
$\%Mg=100-33,8-4,2-45,1=16,9\%$
b. $PTK=180$
$\%C=\dfrac{12.6}{180}.100\%=40\%$
$\%H = \dfrac{12}{180}.100\%=6,7\%$
$\%O=53,3\%$
c. PTK = 141
$\%Na=\dfrac{23.3}{141}.100\%=48,93\%$
$\%O=\dfrac{31}{141}.100\%=21,99\%$
$\%O=29,08\%$
d. $PTK=213$
$\%Al=27:213.100\%=12,67\%$
$\%N=14.2:213.100\%=13,15\%$
$\%O=74,18\%$
e. $PTK=234$
$\%Ca=40:234.100\%=17,09\%$
$\%P=31.2:234.100\%=26,5\%$
$\%O=16.8:234.100\%=54,7\%$
$\%H=1,71\%$
