Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, \(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)
b, Ta có: \(m_{H_2SO_4}=200.9,8\%=19,6\left(g\right)\)
\(\Rightarrow n_{H_2SO_4}=\dfrac{19,6}{98}=0,2\left(mol\right)\)
Theo PT: \(n_{MgO}=n_{MgSO_4}=n_{H_2SO_4}=0,2\left(mol\right)\)
\(\Rightarrow m_{MgO}=0,2.40=8\left(g\right)\)
c, Ta có: m dd sau pư = 8 + 200 = 208 (g)
\(\Rightarrow C\%_{MgSO_4}=\dfrac{0,2.120}{208}.100\%\approx11,54\%\)
\(a.Ba+H_2O\rightarrow Ba\left(OH\right)_2+H_2\\ b.n_{Ba}=\dfrac{1,37}{137}=0,01\left(mol\right)\\ n_{H_2}=n_{Ba}=0,01\left(mol\right)\\ \Rightarrow m_{H_2}=0,01.2=0,02\left(g\right)\\ c.n_{Ba\left(OH\right)_2}=n_{Ba}=0,01\left(mol\right)\\ \Rightarrow m_{Ba\left(OH\right)_2}=0,01.171=1,71\left(g\right)\\ d.m_{ddsaupu}=1,37+72-0,02=73,35\left(g\right)\\ C\%_{Ba\left(OH\right)_2}=\dfrac{1,71}{73,35}.100=2,33\%\)
nZn=0,1 mol
Zn +2HCl=> ZnCl2+ H2
0,1 mol =>0,2 mol
=>mHCl=36,5.0,2=7,3g
=>m dd HCl=7,3/14,6%=50g
mdd sau pứ=6,5+50-0,1.2=56,3g
=>C% dd ZnCl2=(0,1.136)/56,3.100%=24,16%
a.b. Zn + 2HCl ---> ZnCl2 + H2 (1)
Theo pt: 65g 73g 136g 2g
Theo đề: 6,5g 7,3g 13,6g
=> mddHCl=\(\frac{7,3.100}{14,6}=50\left(g\right)\)
c. Từ pt (1), ta có: \(C_{\%}=\frac{13,6}{50+6,5}.100\%=24,1\%\)
\(n_{Na}=\frac{4,6}{23}=0,2\left(mol\right)\)
\(2Na+2H_2O->2NaOH+H_2\) (1)
theo (1) \(n_{NaOH}=n_{Na}=0,2\left(m0l\right)\)
=> \(m_{NaOH}=0,2.40=8\left(g\right)\)
theo (1) \(n_{H_2}=\frac{1}{2}n_{Na}=0,1\left(mol\right)\)
khối lượng dung dịch sau phản ứng là
4,6 + 90 - 0,1.2=94,4(g)
nồng độ % dung dịch thu được là
\(\frac{8}{94,4}.100\%\approx8,47\%\)
Na+H20-> NaOH + 1/2 H2
C%= mNaOH / mNa mH20 - mH2 nhân cho 100
\(Na+H_2O->NaOH+\dfrac{1}{2}H_2\\ a.n_{Na}=\dfrac{m_1}{23}\left(mol\right)\\ m_{ddsau}=\dfrac{m_1}{23}+m_2-\dfrac{m_1}{46}=\dfrac{m_1}{46}+m_2\left(g\right)\\ C\%_B=\dfrac{\dfrac{40}{23}m_1}{\dfrac{m_1}{46}+m_2}\cdot100\%.\\ b.C_M=\dfrac{10dC\%}{M}=10\cdot1,2\cdot\dfrac{0,05}{40}=0,015\left(M\right)\)
\(Na+H_2O->NaOH+\dfrac{1}{2}H_2\\ a.n_{Na}=\dfrac{m_1}{23}\left(mol\right)\\ m_{ddsau}=m_1+m_2-\dfrac{m_1}{23}=\dfrac{22}{23}m_1+m_2\left(g\right)\\ C\%_B=\dfrac{\dfrac{40}{23}m_1}{\dfrac{22}{23}m_1+m_2}\cdot100\%.\\ b.C_M=\dfrac{10dC\%}{M}=10\cdot1,2\cdot\dfrac{0,05}{40}=0,015\left(M\right)\)
\(n_{Na}=\dfrac{2,3}{23}=0,1\left(mol\right)\\ a,PTHH:2Na+H_2SO_4\rightarrow Na_2SO_4+H_2\\ b,n_{Na_2SO_4}=n_{H_2}=n_{H_2SO_4}=\dfrac{0,1}{2}=0,05\left(mol\right)\\ m_{Na_2SO_4}=142.0,05=7,1\left(g\right)\\ c,C\%_{ddH_2SO_4}=\dfrac{0,05.98}{4,9}.100\%=100\%\)
Thường C% < 100% ớ em
a)\(n_K=\dfrac{0,39}{39}=0,01mol\)
\(\left\{{}\begin{matrix}X:KOH\\Y:H_2\end{matrix}\right.\)
b)\(2K+2H_2O\rightarrow2KOH+H_2\)
0,01 0,01 0,01 0,005
\(V_{H_2}=0,005\cdot22,4=0,112l=112ml\)
\(1)2Na+2H_2O\rightarrow2NaOH+H_2\\ 2)n_{Na}=\dfrac{2,3}{23}=0,1mol\\ 2Na+2H_2O\rightarrow2NaOH+H_2\)
0,1 0,1 0,1 0,05
\(m_{NaOH}=0,1.40=4g\\ m_{H_2}=0,05.2=0,1g\\ c)C_{\%NaOH}=\dfrac{4}{2,3+100-0,1}\cdot100=3,91\%\)