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a, PT: \(CaCl_2+2AgNO_3\rightarrow2AgCl_{\downarrow}+Ca\left(NO_3\right)_2\)
b, Ta có: \(n_{CaCl_2}=\dfrac{2,22}{111}=0,02\left(mol\right)\)
\(n_{AgNO_3}=\dfrac{1,7}{170}=0,01\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,02}{1}>\dfrac{0,01}{2}\), ta được CaCl2 dư.
Theo PT: \(n_{AgCl}=n_{AgNO_3}=0,01\left(mol\right)\Rightarrow m_{AgCl}=0,01.143,5=1,435\left(g\right)\)
c, \(n_{CaCl_2\left(pư\right)}=\dfrac{1}{2}n_{AgNO_3}=0,005\left(mol\right)\)
\(\Rightarrow n_{CaCl_2\left(dư\right)}=0,015\left(mol\right)\Rightarrow m_{CaCl_2\left(dư\right)}=0,015.111=1,665\left(g\right)\)
\(a,n_{CaO}=\dfrac{14}{40}=0,35(mol)\\ b,n_{C}=\dfrac{3.10^{-23}}{6.10^{-23}}=0,5(mol)\\ c,n_{H_2O}=\dfrac{9.10^{-23}}{6.10^{-23}}=1,5(mol)\\ d,n_{O_2}=\dfrac{16}{32}=0,5(mol)\)
\(a.n_{H_2}=\dfrac{1,2.10^{23}}{6.10^{23}}=0,2\left(mol\right)\\ n_{SO_2}=\dfrac{6,4}{64}=0,1\left(mol\right)\\ V_X=\left(1,5+2,5+0,2+0,1\right).22,4=96,32\left(l\right)\\b. m_X=1,5.32+2,5.28+0,2.2+6,4=124,8\left(g\right)\)
a.nH2=1,2.10236.1023=0,2(mol)nSO2=6,464=0,1(mol)VX=(1,5+2,5+0,2+0,1).22,4=96,32(l)b.mX=1,5.32+2,5.28+0,2.2+6,4=124,8(g)
a) mN = 0,5 .14 = 7g.
mCl = 0,1 .35.5 = 3.55g
mO = 3.16 = 48g.
b) mN2 = 0,5 .28 = 14g.
mCl2 = 0,1 .71 = 7,1g
mO2 = 3.32 =96g
c) mFe = 0,1 .56 =5,6g mCu = 2,15.64 = 137,6g
mH2SO4 = 0,8.98 = 78,4g.
mCuSO4 = 0,5 .160 = 80g
\(M_{CaCl_2}=40+2.35,5=111\left(\dfrac{g}{mol}\right)\\ m_{CaCl_2}=M_{CaCl_2}.n_{CaCl_2}=111.0,5=55,5\left(g\right)\)