Bài 1: Cho 4,6 gam Na vào 500ml dung dịch CuSO4 1M thu m gam kết tủa và V lít khí ở đktc.Tìm m;V và $C_M$ dung dịch sau phản ứng
Bài 2:Cho 41,1 gam Ba vào 284 gam dung dịch Na2SO4 10% thu m gam kết tủa và V lít khí ở đktc.Tìm m;V và C% dung dịch sau phản ứng
`n_(Na)=\frac{4,6}{23}=0,2(mol)`
`2Na+2H_2O->2NaOH+H_2`
Theo PT
`n_(H_2)=1/2 n_(Na)=0,1(mol)`
`=>V=0,1.22.4=2.24(l)`
`n_(NaOH)=n_(Na)=0,2(mol)`
`n_(CuSO_4)=0,5.1=0,5(mol)`
`CuSO_4+2NaOH->Na_2SO_4+Cu(OH)_2`
Theo PT
`n_(Na_2SO_4)=n_(Cu(OH)_2)=1/2 n_(NaOH)=0,1(mol)`
`=>m_(Cu(OH)_2)=0,1.98=9,8(g)`
`C_(MNa_2SO_4)=\frac{0,1}{0,5}=0,2M`
`C_(MCuSO_4)=\frac{0,4}{0,5}=0,8M`
Bài 2
`n_(Ba)=\frac{41,1}{137}=0,3(mol)`
`n_(Na_2SO_4)=\frac{284.10%}{142}=0,2(mol)`
`Ba+2H_2O->Ba(OH)_2+H_2`
Theo PT
`n_(H_2)=n_(Ba(OH)_2)=n_(Ba)=0,3(mol)`
`=>V_(H_2)=0,3.22,4=6,72(l)`
`Ba(OH)_2+Na_2SO_4->BaSO_4+2NaOH`
Theo PT
`n_(BaSO_4)=n_(Na_2SO_4)=0,2(mol)`
`=>m_(BaSO_4)=0,2.233=46,6(g)`
`m_(dd)=41,1+284-0,2.2-46,6=278,1(g)`
`C%_(Ba(OH)_2)=\frac{0,1.171}{278,1}.100=6,15%`
`C%_(NaOH)=\frac{0,4.40}{278,1}.100=5,75%`
1)
Phản ứng xảy ra:
\(2Na + 2{H_2}O\xrightarrow{{}}2NaOH + {H_2}\)
Ta có:
\({n_{Na}} = \frac{{4,6}}{{23}} = 0,2{\text{ mol = }}{{\text{n}}_{NaOH}};{n_{{H_2}}} = \frac{1}{2}{n_{Na}} = 0,1{\text{ mol}}\)
\( \to V = {V_{{H_2}}} = 0,1.22,4 = 2,24{\text{ lít}}\)
Phản ứng xảy ra:
\(CuS{O_4} + 2NaOH\xrightarrow{{}}Cu{(OH)_2} + N{a_2}S{O_4}\)
Ta có:
\({n_{CuS{O_4}}} = 0,5.1 = 0,5{\text{ mol > }}\frac{1}{2}{n_{NaOH}}\)
Vậy \(CuSO_4\) dư.
\( \to {n_{Cu{{(\^O H)}_2}}} = \frac{1}{2}{n_{NaOH}} = 0,1{\text{ mol}}\)
\( \to m = {m_{Cu{{(OH)}_2}}} = 0,1.(64 + 17.2) = 9,8{\text{ gam}}\)
\( \to {n_{N{a_2}S{O_4}}} = \frac{1}{2}{n_{NaOH}} = 0,1{\text{ mol}}\)
\({n_{CuS{O_4}{\text{ dư}}}} = 0,5 – 0,1 = 0,4{\text{ mol}}\)
\({V_{dd}} = 500{\text{ ml = 0}}{\text{,5 lít}}\)
\( \to {C_{M{\text{ N}}{{\text{a}}_2}S{O_4}}} = \frac{{0,1}}{{0,5}} = 0,2M\)
\({C_{M{\text{ CuS}}{{\text{O}}_4}{\text{ dư}}}} = \frac{{0,4}}{{0,5}} = 0,8M\)
2)
Phản ứng xảy ra:
\(Ba + 2{H_2}O\xrightarrow{{}}Ba{(OH)_2} + {H_2}\)
Ta có:
\({n_{Ba}} = \frac{{41,1}}{{137}} = 0,3{\text{ mol = }}{{\text{n}}_{Ba{{(OH)}_2}}} = {n_{{H_2}}}\)
\( \to V = {V_{{H_2}}} = 0,3.22,4 = 6,72{\text{ lít}}\)
\(N{a_2}S{O_4} + Ba{(OH)_2}\xrightarrow{{}}BaS{O_4} + 2NaOH\)
Ta có:
\({m_{N{a_2}S{O_4}}} = 284.10\% = 28,4{\text{ gam}}\)
\( \to {n_{N{a_2}S{O_4}}} = \frac{{28,4}}{{23.2 + 96}} = 0,2{\text{ mol}}\)
\( \to {n_{Ba{{(OH)}_2}}} > {n_{N{a_2}S{O_4}}} \to N{a_2}S{O_4}\) dư
\( \to {n_{BaS{O_4}}} = {n_{N{a_2}S{O_4}}} = 0,2{\text{ mol}}\)
\( \to m = {m_{BaS{O_4}}} = 0,2.233 = 46,6{\text{ gam}}\)
BTKL:
\({m_{Ba}} + {m_{dd{\text{ N}}{{\text{a}}_2}S{O_4}}} = {m_{dd}} + {m_{{H_2}}} + {m_{BaS{O_4}}}\)
\( \to 41,4 + 284 = {m_{dd}} + 46,6 + 0,3.2 \to {m_{dd}} = 278,2{\text{ gam}}\)
\({n_{NaOH}} = 2{n_{N{a_2}S{O_4}}} = 0,2.2 = 0,4{\text{ mol}}\)
\({n_{Ba{{(OH)}_2}{\text{ dư}}}} = 0,3 – 0,2 = 0,1{\text{mol}}\)
\( \to {m_{NaOH}} = 0,4.40 = 16{\text{ gam}}\)
\({m_{Ba{{(OH)}_2}}} = 0,1.(137 + 17.2) = 17,1{\text{ gam}}\)
\( \to C{\% _{NaOH}} = \frac{{16}}{{278,2}} = 5,75\% \)
\( \to C{\% _{Ba{{(OH)}_2}}} = \frac{{17,1}}{{278,2}} = 6,1466\% \)