Đáp án: C = 70,75%
Giải thích các bước giải:
\(\eqalign{
& {m_{{H_2}S{O_4}}} = {{m{\,_{dd}}_{{H_2}S{O_4}}} \over {100\% }}.C\% = {{100} \over {100\% }}.C\% = c(gam) \cr
& \Rightarrow {m_{{H_2}S{O_4}}} = {{{m_{{H_2}S{O_4}}}} \over {{M_{{H_2}S{O_4}}}}} = {c \over {98}}(mol) \cr
& \Rightarrow {m_{{H_2}O}}\,{(_{trong}}{\,_{{H_2}S{O_4}}}) = m{\,_{dd}}_{{H_2}S{O_4}} - {m_{{H_2}S{O_4}}} = 100 - c\,(gam) \cr
& \Rightarrow {n_{{H_2}O}}\, = {{{m_{{H_2}O}}\,} \over {{M_{{H_2}O}}}} = {{100 - c} \over {18}}\,\,(mol) \cr} \)
giả sử a = 100 gam
=> Khối lượng H2 sinh ra là 0,04694×100 = 4,694 (g) => nH2 = mH2 : MH2 = 4,694 : 2 = 2,347 (mol)
Dựa vào PTHH ta thấy:
\(\eqalign{
& \sum {{n_{{H_2}S{O_4} + {H_2}O}} = \sum {{n_{{H_2}}}} } \cr
& \Rightarrow {c \over {98}} + {{100 - c} \over {18}} = 2,347 \cr
& \Rightarrow 18c + 98 \times (100 - c) = 98 \times 18 \times 2,347 \cr
& \Rightarrow 18c + 9800 - 98c = 4140,108 \cr
& \Rightarrow 80c = 5659,892 \cr
& \Rightarrow c = {{5659,892} \over {80}} \cr
& \Rightarrow c = 70,75 \cr} \)
