Cho 3,6g Mg vào 250g dd Hcl 9,8% a)Thể tích H2? b) C% mỗi chất sau phảng ứng? 18/10/2021 Bởi Jade Cho 3,6g Mg vào 250g dd Hcl 9,8% a)Thể tích H2? b) C% mỗi chất sau phảng ứng?
Đáp án: \({{\text{V}}_{{H_2}}} = 3,36{\text{ lít}}\) \( \to C{\% _{MgC{l_2}}} = 5,626\% ;{\text{ C}}{{\text{\% }}_{HCl{\text{ dư}}}} = 5,35\% \) Giải thích các bước giải: Phản ứng xảy ra: \(Mg + 2HCl\xrightarrow{{}}MgC{l_2} + {H_2}\) Ta có: \({n_{Mg}} = \frac{{3,6}}{{24}} = 0,15{\text{ mol; }}{{\text{m}}_{HCl}} = 250.9,8\% = 24,5{\text{gam}} \to {{\text{n}}_{HCl}} = \frac{{24,5}}{{36,5}} > 2{n_{Mg}}\) do vậy HCl dư. \( \to {n_{{H_2}}} = {n_{Mg}} = 0,15{\text{ mol}} \to {{\text{V}}_{{H_2}}} = 0,15.22,4 = 3,36{\text{ lít}}\) BTKL: \({m_{dd{\text{sau phản ứng}}}} = {m_{Mg}} + {m_{dd{\text{ HCl}}}} – {m_{{H_2}}} = 3,6 + 250 – 0,15.2 = 253,3{\text{ gam}}\) \({n_{HCl{\text{ phản ứng}}}} = 2{n_{{H_2}}} = 0,3{\text{ mol}} \to {{\text{m}}_{HCl{\text{ phản ứng}}}} = 0,3.36,5 = 10,95{\text{gam}}\) \( \to {m_{HCl{\text{ dư}}}} = 24,5 – 10,95 = 13,55{\text{gam; }}{{\text{n}}_{MgC{l_2}}} = {n_{Mg}} = 0,15{\text{ mol}} \to {{\text{m}}_{MgC{l_2}}} = 0,15.(24 + 35,5.2) = 14,25{\text{ gam}}\) \( \to C{\% _{MgC{l_2}}} = \frac{{14,25}}{{253,3}} = 5,626\% ;{\text{ C}}{{\text{\% }}_{HCl{\text{ dư}}}} = \frac{{13,55}}{{253,3}} = 5,35\% \) Bình luận
a, $Mg+ 2HCl \to MgCl_2+ H_2$ $n_{Mg}= \frac{3,6}{24}= 0,15 mol$ $n_{HCl}= \frac{250.9,8\%}{36,5}= 0,67 mol$ => Dư 0,37 mol HCl $n_{H_2}= 0,15 mol \Rightarrow V_{H_2}= 0,15.22,4= 3,36l$ b, $m_{dd spu}= 3,6+250-0,15.2= 253,3g$ $n_{MgCl_2}= 0,15 mol$ $C\%_{MgCl_2}= \frac{95.0,15.100}{253,3}= 5,63\%$ $C\%_{HCl}= \frac{0,37.36,5.100}{253,3}= 5,33\%$ Bình luận
Đáp án:
\({{\text{V}}_{{H_2}}} = 3,36{\text{ lít}}\)
\( \to C{\% _{MgC{l_2}}} = 5,626\% ;{\text{ C}}{{\text{\% }}_{HCl{\text{ dư}}}} = 5,35\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(Mg + 2HCl\xrightarrow{{}}MgC{l_2} + {H_2}\)
Ta có:
\({n_{Mg}} = \frac{{3,6}}{{24}} = 0,15{\text{ mol; }}{{\text{m}}_{HCl}} = 250.9,8\% = 24,5{\text{gam}} \to {{\text{n}}_{HCl}} = \frac{{24,5}}{{36,5}} > 2{n_{Mg}}\)
do vậy HCl dư.
\( \to {n_{{H_2}}} = {n_{Mg}} = 0,15{\text{ mol}} \to {{\text{V}}_{{H_2}}} = 0,15.22,4 = 3,36{\text{ lít}}\)
BTKL:
\({m_{dd{\text{sau phản ứng}}}} = {m_{Mg}} + {m_{dd{\text{ HCl}}}} – {m_{{H_2}}} = 3,6 + 250 – 0,15.2 = 253,3{\text{ gam}}\)
\({n_{HCl{\text{ phản ứng}}}} = 2{n_{{H_2}}} = 0,3{\text{ mol}} \to {{\text{m}}_{HCl{\text{ phản ứng}}}} = 0,3.36,5 = 10,95{\text{gam}}\)
\( \to {m_{HCl{\text{ dư}}}} = 24,5 – 10,95 = 13,55{\text{gam; }}{{\text{n}}_{MgC{l_2}}} = {n_{Mg}} = 0,15{\text{ mol}} \to {{\text{m}}_{MgC{l_2}}} = 0,15.(24 + 35,5.2) = 14,25{\text{ gam}}\)
\( \to C{\% _{MgC{l_2}}} = \frac{{14,25}}{{253,3}} = 5,626\% ;{\text{ C}}{{\text{\% }}_{HCl{\text{ dư}}}} = \frac{{13,55}}{{253,3}} = 5,35\% \)
a, $Mg+ 2HCl \to MgCl_2+ H_2$
$n_{Mg}= \frac{3,6}{24}= 0,15 mol$
$n_{HCl}= \frac{250.9,8\%}{36,5}= 0,67 mol$
=> Dư 0,37 mol HCl
$n_{H_2}= 0,15 mol \Rightarrow V_{H_2}= 0,15.22,4= 3,36l$
b,
$m_{dd spu}= 3,6+250-0,15.2= 253,3g$
$n_{MgCl_2}= 0,15 mol$
$C\%_{MgCl_2}= \frac{95.0,15.100}{253,3}= 5,63\%$
$C\%_{HCl}= \frac{0,37.36,5.100}{253,3}= 5,33\%$