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Goi $m_{dd\ H_2SO_4} = 139(gam)$
Ta có :
$V_{dd} = m : D = 139 : 1,39 = 100(ml) = 0,1(lít)$
$n_{H_2SO_4} = 0,1.6,95 = 0,695(mol)$
Vậy :
$C\%_{H_2SO_4} = \dfrac{0,695.98}{139}.100\% = 49\%$
Khối lượng dd NaOH : 1,28*250= 80g => nNaOH = (80*25/100)/40 = 2mol
PT : BaCl2 + H2SO4 ----> BaSO4 + 2HCl
0,05mol --> 0,05mol
H2SO4 + 2NaOH -------> Na2SO4 + 2H2O
1mol <--- 2mol
hoep t trên ta có tổng số mol của H2SO4 : 0,05+1 = 1,05mol => mH2SO4 = 98*1,05 =102,9g
Vậy c%H2SO4 : 102,9/200*100= 51,45%
\(a) n_{Zn} = \dfrac{19,5}{65} = 0,3(mol) ; n_{HCl} = 0,35.2 = 0,7(mol)\\ Zn + 2HCl \to ZnCl_2 + H_2\\ n_{HCl} = 0,7 > 2n_{Zn} = 0,6 \to HCl\ dư\\ n_{H_2} = n_{Zn} = 0,3(mol) \Rightarrow V_{H_2} = 0,3.22,4 = 6,72(lít)\\ b) m_{dd\ HCl} = 350.1,05 = 367,5(gam)\\ m_{dd\ sau\ pư} = 19,5 + 367,5 - 0,3.2 = 386,4(gam)\\ \Rightarrow C\%_{ZnCl_2} = \dfrac{0,3.136}{386,4}.100\% = 10,56\%\\ c) C\%_{HCl} = \dfrac{0,7.36,5}{367,5}.100\% = 6,95\%\)
Bài 1 :
Giả sử thể tích dung dịch H2SO4 là V ml
\(\rightarrow m_{dd}=1,84V\left(g\right)\rightarrow m_{H2SO4}=1,84V.98\%=1,8032\left(V\right)\)
\(\rightarrow n_{H2SO4}=\frac{1,8032V}{98}=0,0184V\left(mol\right)\)
\(\rightarrow CM_{H2SO4}=\frac{0,0184V.1000}{V}=18,4M\)
\(n_{H2SO4}=2.2,5=5\left(mol\right)\rightarrow m_{H2SO4}=5.98=490\left(g\right)\)
\(\rightarrow m_{dd_{H2SO4_{Can}}}=\frac{490}{98\%}=500\left(g\right)\)
Vậy V dung dịch H2SO4 cần \(=\frac{500}{1,84}=271,74\left(ml\right)\)
Cho 271,74 ml H2SO4 98% vào dung dịch, sau đó thêm H2O vào đủ 2 lít/
Bài 2:
Gọi số mol Na2O cần là x \(\rightarrow m_{Na2O}=62x\)
\(\rightarrow\) m dung dịch sau khi thêm=62x+84,5 gam
\(Na_2O+H_2O\rightarrow2NaOH\)
\(\rightarrow n_{NaOH_{tao.ra}}=2x\rightarrow m_{NaOH_{tao.ra}}=2x.40=80x\left(g\right)\)
\(\rightarrow\) m NaOH trong dung dịch \(=80x+84,5.10\%=80x+8,45\left(g\right)\)
\(\rightarrow C\%_{NaOH}=\frac{\left(80x+8,45\right)}{\left(62x+84,5\right)}=28,45\%\rightarrow x=0,25\)
\(\rightarrow m_{Na2O}=15,5\left(g\right)\)
Bài 3 :
\(n_{MgCO3}=\frac{16,8}{84}=0,2\left(mol\right)\)
\(n_{HCl}=\frac{200.10,95\%}{36,5}=0,6\left(mol\right)\)
\(MgCO_3+2HCl\rightarrow MgCl_2+CO_2+H_2O\)
Nên HCl dư
\(n_{CO2}=0,2\left(mol\right)\)
\(n_{HCl_{du}}=0,6-0,2.2=0,2\left(mol\right)\)
\(m_{dd_{Spu}}=16,8+200-0,2.44=208\left(g\right)\)
\(C\%_{HCl}=\frac{0,2.36,5}{208}.100\%=3,51\%\)
\(C\%_{MgCl2}=\frac{0,2.95}{208}.100\%=9,13\%\)
\(n_{Ba}=\dfrac{6,85}{137}=0,05\left(mol\right)\\ m_{H_2SO_4}=500.1,96\%=9,8\left(g\right)\\ PTHH:Ba+H_2SO_4\rightarrow BaSO_4+H_2\uparrow\\ LTL:0,05< 0,1\Rightarrow H_2SO_4.dư\)
\(n_{BaSO_4}=n_{H_2SO_4\left(pư\right)}=n_{Ba}=n_{H_2}=0,05\left(mol\right)\)
\(\Rightarrow n_{H_2SO_4\left(dư\right)}=0,1-0,05=0,05\left(mol\right)\)
\(V_{dd}=\dfrac{500}{1,15}\approx434\left(ml\right)=0,434\left(l\right)\)
\(C_{MBaSO_4}=\dfrac{0,05}{0,434}=0,115M\\ C_{MH_2SO_4\left(dư\right)}=\dfrac{0,05}{0,434}=0,115M\)
PTHH: \(K+H_2O\rightarrow KOH+\dfrac{1}{2}H_2\uparrow\)
a) Ta có: \(n_K=\dfrac{7,8}{39}=0,2\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{KOH}=0,2\left(mol\right)\\n_{H_2}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{KOH}=0,2\cdot56=11,2\left(g\right)\\m_{H_2}=0,1\cdot2=0,2 \left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd\left(saup/ứ\right)}=m_K+m_{H_2O}-m_{H_2}=400\left(g\right)\)
\(\Rightarrow C\%_{KOH}=\dfrac{11,2}{400}\cdot100\%=2,8\%\)
b) Ta có: \(V_{dd\left(saup/ứ\right)}=\dfrac{400}{1,08}\approx370,37\left(ml\right)=0,37037\left(l\right)\)
\(\Rightarrow C_{M_{KOH}}=\dfrac{0,2}{0,37037}\approx0,54\left(M\right)\)
\(n_{H_2SO_4}=0.1\cdot2=0.2\left(mol\right)\)
\(m_{dd_{H_2SO_4}}=100\cdot1.2=120\left(g\right)\)
\(n_{BaCl_2}=0.1\cdot1=0.1\left(mol\right)\)
\(m_{dd_{BaCl_2}}=100\cdot1.32=132\left(g\right)\)
\(BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl\)
\(0.1................0.1.........0.1...............0.2\)
\(\Rightarrow H_2SO_4dư\)
\(m_{BaSO_4}=0.1\cdot233=23.3\left(g\right)\)
\(V_{dd}=0.1+0.1=0.2\left(l\right)\)
\(C_{M_{H_2SO_4\left(dư\right)}}=\dfrac{0.2-0.1}{0.2}=0.5\left(M\right)\)
\(C_{M_{HCl}}=\dfrac{0.2}{0.2}=1\left(M\right)\)
\(m_{\text{dung dịch sau phản ứng}}=120+132-23.3=228.7\left(g\right)\)
\(C\%_{H_2SO_4\left(dư\right)}=\dfrac{0.1\cdot98}{228.7}\cdot100\%=4.28\%\)
\(C\%_{HCl}=\dfrac{0.2\cdot36.5}{228.7}\cdot100\%=3.2\%\)
Xem lại đề nhé
à phải là g/ml