Cho 3,6g Mg vào 250g dd HCl 9,8% a/VH2=? b/C% mỗi chất sau phản ứng 18/10/2021 Bởi Genesis Cho 3,6g Mg vào 250g dd HCl 9,8% a/VH2=? b/C% mỗi chất sau phản ứng
a,mHCl=$\frac{9,8*250}{100}$=24.5(g)=>nHCl=$\frac{24.5}{36.5}$=0.67(mol) mMg=3.6(g)=>nMg=$\frac{3.6}{24}$=0.15(mol) ta có phương trình:Mg+2HCl=>MgCl2+H2 ta có tỉ lệ:nMg:nHCl=$\frac{0.15}{1}$<$\frac{0.67}{2}$(Mg hết,HCl dư) =>nH2=0.15(mol)=>vH2=0.15*22.4=3.36(lít):nHCl=0.15*2=0.3(mol)=>mHCl=0.3*36.5=10.95(g) b,nMgCl2=0.15(mol)=>mMgCl2=0.15*(24+35.5*2)=14.25(g) C%MgCl2=$\frac{14.25}{3.6+10.95}$*100=97.9% 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: \({n_{Mg}} = \frac{{3,6}}{{24}} = 0,15{\text{ mol; }}{{\text{m}}_{HCl}} = 250.98\% = 24,5{\text{gam}} \to {{\text{n}}_{HCl}} = \frac{{24,5}}{{36,5}} > 2{n_{Mg}}\) 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,mHCl=$\frac{9,8*250}{100}$=24.5(g)=>nHCl=$\frac{24.5}{36.5}$=0.67(mol)
mMg=3.6(g)=>nMg=$\frac{3.6}{24}$=0.15(mol)
ta có phương trình:Mg+2HCl=>MgCl2+H2
ta có tỉ lệ:nMg:nHCl=$\frac{0.15}{1}$<$\frac{0.67}{2}$(Mg hết,HCl dư)
=>nH2=0.15(mol)=>vH2=0.15*22.4=3.36(lít):nHCl=0.15*2=0.3(mol)=>mHCl=0.3*36.5=10.95(g)
b,nMgCl2=0.15(mol)=>mMgCl2=0.15*(24+35.5*2)=14.25(g)
C%MgCl2=$\frac{14.25}{3.6+10.95}$*100=97.9%
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:
\({n_{Mg}} = \frac{{3,6}}{{24}} = 0,15{\text{ mol; }}{{\text{m}}_{HCl}} = 250.98\% = 24,5{\text{gam}} \to {{\text{n}}_{HCl}} = \frac{{24,5}}{{36,5}} > 2{n_{Mg}}\)
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:
\({m_{dd{\text{sau phản ứng}}}} = {m_{Mg}} + {m_{dd{\text{ HCl}}}} – {m_{{H_2}}} = 3,6 + 250 – 0,15.2 = 253,3{\text{ gam}}\)
\({n_{HCl{\text{ phản ứng}}}} = 2{n_{{H_2}}} = 0,3{\text{ mol}} \to {{\text{m}}_{HCl{\text{ phản ứng}}}} = 0,3.36,5 = 10,95{\text{gam}}\)
\( \to {m_{HCl{\text{ dư}}}} = 24,5 – 10,95 = 13,55{\text{gam; }}{{\text{n}}_{MgC{l_2}}} = {n_{Mg}} = 0,15{\text{ mol}} \to {{\text{m}}_{MgC{l_2}}} = 0,15.(24 + 35,5.2) = 14,25{\text{ gam}}\)
\( \to C{\% _{MgC{l_2}}} = \frac{{14,25}}{{253,3}} = 5,626\% ;{\text{ C}}{{\text{\% }}_{HCl{\text{ dư}}}} = \frac{{13,55}}{{253,3}} = 5,35\% \)