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Câu 1:
\(n_{H2}=0,17\left(mol\right)\)
\(n_{HCl}=\frac{300.10,22}{100.36,5}=0,84\left(mol\right)>2n_{H2}=0,34\left(mol\right)\) nên HCl dư
\(\Rightarrow n_{HCl_{pư}}=2n_{H2}=2.0,17=0,34\left(mol\right)\)
Áp dụng ĐLBT Khối Lượng , ta có :
mX + mHCL = m muối + mH2
\(\Leftrightarrow m_{muoi}=6,1+0,34.36,5-0,34=18,17\left(g\right)\)
Câu 2:
\(n_{HCl}=\frac{400.7,3\%}{36,5}=0,8\left(mol\right)\)
\(n_{MgO}=\frac{2,4}{40}=0,06\left(mol\right)\)
\(n_{MgCO3}=\frac{13,6}{84}=0,15\left(mol\right)\)
\(n_{FeCO3}=\frac{10,44}{116}=0,09\left(mol\right)\)
\(n_{MgCl2}=n_{MgCO3}+n_{MgO}=0,21\)
\(n_{FeCl2}=n_{FeCO3}=0,09\left(mol\right)\)
\(n_{HCl\left(Pư\right)}=2n_{FeCl2}+2n_{MgCl2}=0,6\left(mol\right)\)
\(\Rightarrow n_{HCl\left(dư\right)}=0,8-0,6=0,2\left(mol\right)\)
\(n_{CO2}=n_{MgCO3}+n_{FeCO3}=0,24\left(mol\right)\)
\(m_{dd\left(spu\right)}=400+12,6+10,44+2,4-0,24.44=414,88\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{FeCl2}=\frac{0,09.127}{414,88}.100\%=2,755\%\\C\%_{MgCl2}=\frac{0,21.95}{414,88}.100\%=4,81\%\\C\%_{HCl\left(dư\right)}=\frac{0,2.36,5}{414,88}.100\%=1,76\%\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}n_{Fe\left(OH\right)2}=n_{FeCl2}=0,09\\n_{Mg\left(OH\right)2}=n_{MgCl2}=0,21\end{matrix}\right.\)
\(\Rightarrow m_{kt}=m_{Mg\left(OH\right)2}+m_{Fe\left(OH\right)2}\)
\(\Leftrightarrow m_{kt}=0,21.58+0,09.90=20,28\left(g\right)\)
Câu 1:
\(27n_{Al}+56n_{Fe}=8,715\)
\(n_{Al}-n_{Fe}=0\)
\(\Rightarrow n_{Al}=n_{Fe}=0,105\)
\(n_{Al}=n_{AlCl3}=0,105\left(mol\right)\)
\(\Rightarrow m_{muoi}=m_{AlCl3}+m_{FeCl3}\)
\(\Leftrightarrow m_{muoi}=0,105.162,5+0,105+133,5=31,08\left(g\right)\)
Câu 2: Mai mình làm cho giờ muộn rồi lười gõ telexx
Câu 1:
\(n_{KBr}=\frac{19,04}{119}=0,16\left(mol\right)\)
\(n_{NaBr}=\frac{12,36}{103}=0,12\left(mol\right)\)
\(n_{NaI}=\frac{13,5}{150}=0,09\left(mol\right)\)
PTHH :
\(2KBr+Cl_2\underrightarrow{^{to}}2KCl+Br_2\)
0,16____0,08 ___0,16_______(mol)
\(2NaBr+Cl_2\underrightarrow{^{to}}2NaCl+Br_2\)
0,12____0,06 ____ 0,12 _____(mol)
\(2NaI+Cl_2\underrightarrow{^{to}}2NaCl+I_2\)
0,09 ___0,045 ___0,09 ____ (mol)
\(n_{Cl2}=0,08+0,06+0,045=0,185\left(mol\right)\)
\(\Rightarrow V_{Cl2}=4,144\left(l\right)\)
\(\left\{{}\begin{matrix}n_{KCl}=0,16\left(mol\right)\\n_{NaCl}=0,12+0,09=0,21\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow m_{muoi}=m_{KCl}+m_{NaCl}=0,16+74,5+0,21.58,5=24,205\left(g\right)\)
Câu 2:Gọi CTTQ của hh X là R2On
\(n_{HCl}=0,24.2=0,48\left(mol\right)\)
\(PTHH:R_2O_n+2nHCl\rightarrow2RCl_n+nH_2O\)
Theo PT: \(n_{H2O}=\frac{1}{2}n_{HCl}=0,24\left(mol\right)\)
Áp dụng định luật BT khối lượng : mX + mHCl = m muối + mH2O
\(\Rightarrow m_{muoi}=9,125+0,48.36,5-0,24.18=22,325\left(g\right)\)
Câu 3:
\(n_{NaCl}=0,05.0,4=0,02\left(mol\right)\)
\(n_{NaI}=0,4.0,15=0,06\left(mol\right)\)
\(AgNO_3+NaCl\rightarrow AgCl+NaNO_3\)
\(AgNO_3+NaI\rightarrow AgI+NaNO_3\)
\(n_{AgCl}=n_{NaCl}=0,02\left(mol\right)\)
\(\Rightarrow n_{AgI}=n_{NaI}=0,06\left(mol\right)\)
\(\Rightarrow m=m_{AgCl}+m_{AgI}=0,02.143,5+0,06.235=16,97\left(g\right)\)
