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a. Mg + 2HCl ZnCl2 + H2
0,05 mol 0,1 mol 0,05 mol
b. mMg =0,05.24 = 1,2 gam
mHClbanđầu = mHClpu + mHCl dư
= 3,65 + 3,65.20% = 4,38gam
1. \(6NaOH+Fe_2\left(SO_4\right)_3\rightarrow2Fe\left(OH\right)_3+3Na_2SO_4\)
2. \(Mg+2AgNO_3\rightarrow Mg\left(NO_3\right)_2+2Ag\)
3.\(4Na+O_2\rightarrow2Na_2O\)
4.\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
chúc bn học tốt
1. 6NaOH + Fe2(SO4)3 ➝ 2Fe(OH)3 + 3Na2SO4
2. Mg + 2AgNO3 ➝ Mg(NO3)2 + 2Ag
3. 4Na + O2 \(\underrightarrow{t^o}\) 2Na2O
4. 2Al + 6HCl \(\rightarrow\) 2AlCl3 + 3H2
\(n_{Cu\left(NO_3\right)_2}=\dfrac{15.04}{188}=0.08\left(mol\right)\)
\(n_{O_2}=x\left(mol\right)\)
\(2Cu\left(NO_3\right)_2\underrightarrow{t^0}2CuO+4NO_2+O_2\)
\(2x......................4x......x\)
\(BTKL:\)
\(m_{NO_2}+m_{O_2}=15.04-8.56=6.48\left(g\right)\)
\(\Rightarrow4x\cdot46+32x=6.48\)
\(\Rightarrow x=0.03\)
\(\%Cu\left(NO_3\right)_{2\left(ph\right)}=\dfrac{0.03}{0.08}\cdot100\%=37.5\%\)
\(b.\)
\(\overline{M}=\dfrac{6.48}{0.12+0.03}=43.2\left(\dfrac{g}{mol}\right)\)
\(d_{\dfrac{hh}{H_2}}=\dfrac{43.2}{2}=21.6\)
\(c.\)
\(H\%=\dfrac{0.03}{0.08}\cdot100\%=37.5\%\)
\(\)
\(PTHH:Mg+H_2SO_4--->MgSO_4+H_2\uparrow\)
Áp dụng ĐLBTKL, ta có:
\(m_{Mg}+m_{H_2SO_4}=m_{MgSO_4}+m_{H_2}\)
\(\Leftrightarrow9,6+39,2=m_{MgSO_4}+0,8\)
\(\Leftrightarrow m_{MgSO_4}=9,6+39,2-0,8=48\left(g\right)\)
2.
a; MgCO3 -> MgO + CO2
BaCO3 -> BaO + CO2
b; 2NaNO3 -> 2NaNO2 + O2
2KNO3 -> 2KNO2 + O2
c; 2Mg(NO3)2 -> 2MgO + 4NO2 + O2
2Cu(NO3)2 -> 2CuO + 4NO2 + O2
2Pb(NO3)2 -> 2PbO + 4NO2 + O2
2.
a; MgCO -> MgO + CO BaCO -> BaO + CO
b; 2NaNO -> 2NaNO + O 2KNO
-> 2KNO + O
c; 2Mg(NO ) -> 2MgO + 4NO + O 2Cu(NO )
-> 2CuO + 4NO + O 2Pb(NO ) -> 2PbO + 4NO + O
tick cho mik nha
a) Mg + 2HCl --> MgCl2 + H2
b) \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: Mg + 2HCl --> MgCl2 + H2
0,3<--0,15<--0,15
=> mHCl = 0,3.36,5 = 10,95 (g)
c) mMgCl2 = 0,15.95 = 14,25 (g)
a) Mg + 2AgNO3 → Mg(NO3)2 + 2Ag↓
b) \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
Theo PT: \(n_{AgNO_3}=2n_{Mg}=2\times0,2=0,4\left(mol\right)\)
\(\Rightarrow m_{AgNO_3}=0,4\times170=68\left(g\right)\)
Theo PT: \(n_{Mg\left(NO_3\right)_2}=n_{Mg}=0,2\left(mol\right)\)
\(\Rightarrow m_{Mg\left(NO_3\right)_2}=0,2\times148=29,6\left(g\right)\)