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\(a) n_{Zn(NO_3)_2} = \dfrac{37,8}{189} = 0,2(mol)\\ n_{Zn} = 0,2\ mol \to m_{Zn} = 0,2.65 = 13\ gam\\ n_N = 0,2.2 = 0,4\ mol \to m_N = 0,4.14 = 5,6\ gam\\ m_O = 37,5 - 13 - 5,6 = 18,9(gam)\\ b)n_{Fe_3(PO_4)_2} = \dfrac{10,74}{358} = 0,03(moL)\\ n_{Fe} = 0,03.3 = 0,09 \to m_{Fe} = 0,09.56 = 5,04(gam)\\ n_P = 0,03.2 = 0,06 \to m_P = 0,06.31 = 1,86(gam)\\ m_O = 10,74 - 5,04 -1,86 = 3,84(gam)\\ c) n_{Al} = 0,2.2 = 0,4(mol\to m_{Al} = 0,4.27 = 10,8(gam)\\ n_S = 0,2.3 = 0,6 \to m_S = 0,6.32 = 19,2(gam)\\ n_O = 0,2.12 = 2,4 \to m_O = 2,4.16 = 38,4(gam)\)
\(d) n_{Zn(NO_3)_2} = \dfrac{6.10^{20}}{6.10^{23}} = 0,001(mol)\\ n_{Zn} = 0,001 \to m_{Zn} = 0,001.65 = 0,065(gam)\\ n_N = 0,001.2 = 0,002 \to m_N = 0,002.14 = 0,028(gam)\\ n_O = 0,001.6 = 0,006 \to m_O = 0,006.16= 0,096(gam)\)
Theo gt ta có: $n_{Zn(NO_3)_2}=0,2(mol);n_{Fe_3(PO_4)_2}=0,03(mol);n_{Zn(NO_3)_2}=1(mol)$
a, $m_{Zn}=13(g);m_{N}=5,6(g);m_{O}=19,2(g)$
b, $m_{Fe}=5,04(g);m_{P}=1,86(g)$;m_{O}=3,84(g)$
c, $m_{Al}=10,8(g);m_{S}=19,2(g);m_{O}=38,4(g)$
d, $m_{Zn}=65(g);m_{N}=28(g);m_{O}=96(g)$
a, nBaCl2 = 26 : (137+2.35,5) = 0,125 (mol)
Cứ 1mol BaCl2 có 1mol Ba, 2mol Cl
\(\rightarrow\) 0,125mol BaCl2 có 0,125mol Ba, 0,25mol Cl
mBa = 0,125.137 = 17,125 (g)
mCl = 26-17,125 = 8,875 (g)
b, nMnCl2 = 7,56 : (55+2.35,5) = 0,06 (mol)
Cứ 1mol MnCl2 có 1mol Mn, 2mol Cl
\(\rightarrow\) 0,06mol MnCl2 có 0,06mol Mn, 0,12mol Cl
mMn = 0,06.55 = 3,3 (g)
mCl = 7,56-3,3 = 4,26 (g)
c, nAgNO3 = 105,4 : (108+14+3.16) = 0,62 (mol)
Cứ 1mol AgNO3 có 1mol Ag, 1mol N, 3mol O
\(\rightarrow\) 0,62mol AgNO3 có 0,62mol Ag, 0,62mol N, 1,86mol O
mAg= 0,62.108 = 66,96 (g)
mN = 0,62.14 = 8,68 (g)
mO = 105,4-66,96-8,68 = 29,76 (g)
d, nZn(NO3)3 = 37,8 : (65+3.14+9.16) ~ 0,15 (mol)
Cứ 1mol Zn(NO3)3 có 1mol Zn, 3mol N, 9mol O
\(\rightarrow\) 0,15mol Zn(NO3)3 có 0,15mol Zn, 0,45mol N, 1,35mol O
mZn= 0,15.65 = 9,75 (g)
mN = 0,45.14 = 6,3 (g)
mO = 37,8-9,75-6,3 = 21,75 (g)
e, nAl2(SO4)3 = 34,2 : (2.27+3.32+12.16) = 0,1 (mol)
Cứ 1mol Al2(SO4)3 có 2mol Al, 3mol S, 12mol O
\(\rightarrow\) 0,1mol Al2(SO4)3 có 0,2mol Al, 0,3mol S, 1,2mol O
mAl = 0,2.27 = 5, 4(g)
mS = 0,3.32 = 9,6 (g)
mO = 34,2-5,4-9,6 = 19,2 (g)
CHÚC BẠN HỌC TỐT ! ^^
a, \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
b, \(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)
c, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
d, \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
e, \(2NaOH+CuSO_4\rightarrow Na_2SO_4+Cu\left(OH\right)_2\)
f, \(Al_2O_3+2NaOH\rightarrow2NaAlO_2+H_2O\)
g, \(2Fe\left(OH\right)_3\underrightarrow{t^o}Fe_2O_3+3H_2O\)
h, \(2H_3PO_4+3Ca\left(OH\right)_2\rightarrow Ca_3\left(PO_4\right)_2+6H_2O\)
i, \(BaCl_2+2AgNO_3\rightarrow2AgCl+Ba\left(NO_3\right)_2\)
k, \(2FeO+4H_2SO_{4đ}\rightarrow Fe_2\left(SO_4\right)_3+SO_2+4H_2O\)
Bạn tham khảo nhé!
a)
MnO2: Mn+4, O-2
HCl: H+1, Cl-1
HClO4: H+1, Cl+7, O-2
MnCl2: Mn+2, Cl-1
MnSO4: Mn+2, S+6, O-2
b)
CaCO3: Ca+2, C+4, O-2
HNO3: H+1, N+5, O-2
HClO4: H+1, Cl+7, O-2
BaCl2: Ba+2, Cl-1
BaSO4: Ba+2, S+6, O-2
c)
NH4+: N-3, H+1
SO42-: S+6, O-2
NO3-: N+5, O-2
CO32-: C+4, O-2
PO43-: P+5, O-2
\(a,MnO_2\left\{{}\begin{matrix}Mn^{4+}\\O^{2-}\end{matrix}\right.\\ HCl\left\{{}\begin{matrix}H^+\\Cl^-\end{matrix}\right.\\ HClO_4\left\{{}\begin{matrix}H^+\\ClO_7^-\end{matrix}\right.\\ MnCl_2\left\{{}\begin{matrix}Mn^{2+}\\Cl^-\end{matrix}\right.\\ MnSO_4\left\{{}\begin{matrix}Mn^{2+}\\SO_4^{2-}\end{matrix}\right.\\ b,CaCO_3\left\{{}\begin{matrix}Ca^{2+}\\CO_3^{2-}\end{matrix}\right.\)
\(HNO_3\left\{{}\begin{matrix}H^+\\NO_3^-\end{matrix}\right.\\ HClO_4\left(đã.làm\right)\\ BaCl_2\left\{{}\begin{matrix}Ba^{2+}\\Cl^-\end{matrix}\right.\\ BaSO_4\left\{{}\begin{matrix}Ba^{2+}\\SO_4^{2-}\end{matrix}\right.\\ c,NH_4^+\\ SO_4^{2-}\\ NO_3^-\\ CO_3^{2-}\\ PO_4^{3-}\)
a) nBaCl2 = \(\dfrac{26}{208}\)= 0,125 (mol) = nBa
⇒mBa = 0,125 . 137 = 17,125 g
⇒mCl2 = 26 - 17,125 = 8,875 g