Đáp án:
Bạn tham khảo lời giải ở dưới nhé!!!
Giải thích các bước giải:
Bài 1:
\(\begin{array}{l}
{H_2}S{O_4}\\
M = 98\\
\% {m_H} = \dfrac{2}{{98}} \times 100\% = 2,04\% \\
\% {m_S} = \dfrac{{32}}{{98}} \times 100\% = 32,65\% \\
\% {m_O} = \dfrac{{16 \times 4}}{{98}} \times 100\% = 65,31\% \\
F{e_2}{(C{O_3})_3}\\
M = 292\\
\% {m_{Fe}} = \dfrac{{56 \times 2}}{{292}} \times 100\% = 38,36\% \\
\% {m_C} = \dfrac{{12 \times 3}}{{292}} \times 100\% = 12,33\% \\
\% {m_O} = \dfrac{{16 \times 9}}{{292}} \times 100\% = 49,31\% \\
Zn{(OH)_2}\\
M = 99\\
\% {m_{Zn}} = \dfrac{{65}}{{99}} \times 100\% = 65,66\% \\
\% {m_O} = \dfrac{{16 \times 2}}{{99}} \times 100\% = 32,32\% \\
\% {m_H} = \dfrac{2}{{99}} \times 100\% = 2,02\% \\
AgN{O_3}\\
M = 170\\
\% {m_{Ag}} = \dfrac{{108}}{{170}} \times 100\% = 63,53\% \\
\% {m_N} = \dfrac{{14}}{{170}} \times 100\% = 8,24\% \\
\% {m_O} = \dfrac{{16 \times 3}}{{170}} \times 100\% = 28,23\% \\
Al{(N{O_3})_3}\\
M = 213\\
\% {m_{Al}} = \dfrac{{27}}{{213}} \times 100\% = 12,68\% \\
\% {m_N} = \dfrac{{14 \times 3}}{{213}} \times 100\% = 19,71\% \\
\% {m_O} = \dfrac{{16 \times 9}}{{213}} \times 100\% = 67,61\%
\end{array}\)
