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\(n_{CO_2}=0,2\left(mol\right)\)
a)\(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\) (1)
b)Từ (1)\(\Rightarrow n_{CaCO_3}=n_{CO_2}=0,2mol\)
\(\Rightarrow m_{CaCO_3}=100.0,2=20\left(g\right)\)
c)Từ (1)\(\Rightarrow n_{Ca\left(OH\right)_2}=n_{CO_2}=0,2mol\)
\(CaO+H_2O\rightarrow Ca\left(OH\right)_2\)
\(0,2\) \(0,2\) (mol)
\(\Rightarrow n_{CaO}=0,2mol\)
\(\Rightarrow m_{CaO}=11,2\left(g\right)\)
a, \(CaO+Ca\left(OH\right)_2->CaCO_3+H_2O\)
b,\(n_{CaCO_3}=n_{CO_2}=0,2mol\) \(->m_{CaCO_3}=0,2.100=20g\)
c,\(CaO+H_2O->Ca\left(OH\right)_2\)
\(n_{CaO}=n_{Ca\left(OH\right)_2}=0,2mol\) \(->m_{CaO}=56.0,2=11,2g\)
a) Ta có: \(n_{Ca\left(OH\right)_2}=\dfrac{14,8}{74}=0,2\left(mol\right)\) \(\Rightarrow V_{ddCa\left(OH\right)_2}=\dfrac{0,2}{0,2}=1\left(l\right)\)
b) Ta có: \(n_{Pb\left(NO_3\right)_2}=\dfrac{6,62}{331}=0,02\left(mol\right)\) \(\Rightarrow V_{ddPb\left(NO_3\right)_2}=\dfrac{0,02}{0,1}=0,2\left(l\right)\)
a) PTHH: CaO + H2O ===> Ca(OH)2
b) nCaO = 2,8 / 56 = 0,05 (mol)
=> nCa(OH)2 = nCaO = 0,05 (mol)
=> mCa(OH)2 = 0,05 x 74 = 3,7 (gam)
a) Ta có:
nCaO= \(\frac{m_{CaO}}{M_{CaO}}=\frac{2,8}{56}=0,05\left(mol\right)\)
PTHH: CaO + H2O -> Ca(OH)2
b) Theo PTHH và đề bài, ta có:
\(n_{Ca\left(OH\right)_2}=n_{CaO}=0,05\left(mol\right)\)
Khối lượng Ca(OH)2 thu được:
\(m_{Ca\left(OH\right)_2}=n_{Ca\left(OH\right)_2}.M_{Ca\left(OH\right)_2}=0,05.74=3,7\left(g\right)\)
a)
\(n_{CaO}=\dfrac{2,8}{56}=0,05\left(mol\right)\)
PTHH: CaO + H2O --> Ca(OH)2
0,05------------>0,05
=> mCa(OH)2 = 0,05.74 = 3,7 (g)
b) dH2O = 1 g/ml
=> mH2O = 400.1 = 400 (g)
mdd = 2,8 + 400 = 402,8 (g)
\(m_{CO_2}=13,4-6,8=6,6\left(g\right)\)
=> \(n_{CO_2}=\dfrac{6,6}{44}=0,15\left(mol\right)\)
=> \(n_{CaCO_3}=0,15\left(mol\right)\)
=> \(m_{CaCO_3}=0,15.100=15\left(g\right)\)
nCH4 = 11.2/22.4 = 0.5 (mol)
CH4 + 2O2 -to-> CO2 + 2H2O
0.5____________0.5
CO2 + Ca(OH)2 => CaCO3 + H2O
0.5_______________0.5
mCaCO3 = 0.5*100 = 50 (g)
\(n_{CaO}=\dfrac{8,4}{56}=0,15mol\)
\(m_{H_2O}=\dfrac{1,8}{18}=0,1mol\)
\(CaO+H_2O\rightarrow Ca\left(OH\right)_2\)
0,15 > 0,1 ( mol )
0,1 0,1 ( mol )
\(m_{Ca\left(OH\right)_2}=0,1.74=7,4g\)
\(n_{CaO}=\dfrac{8,4}{56}=0,15\left(mol\right)\)
\(n_{H2O}=\dfrac{1,8}{18}=0,1\left(mol\right)\)
\(pthh:CaO+H_2O->Ca\left(OH\right)_2\)
LTL : \(\dfrac{0,15}{1}>\dfrac{0,1}{1}\)
=> CaO dư , H2O hết
\(theopthh:n_{Ca\left(OH\right)_2}=n_{H_2O}=0,1\left(mol\right)\)
=>m= \(m_{Ca\left(OH\right)_2}=0,1.74=7,4\left(G\right)\)