a) Ca(OH)2 + CO2 \(\rightarrow\) CaCO3 + H2O
b) Ta có: nCO2 =\(\frac{5.6}{22.4}\)= 0.25 (mol)
Theo ptpư, ta có: nCaCO3 = nCO2 = 0.25 (mol)
=> mCaCO3 = 0.25 \(\times\) 100 = 25 (g)
c) Theo ptpư, ta có: nCa(OH)2 = nCO2 = 0.25 (mol)
=> Cm Ca(OH)2 = \(\frac{0.25}{0.1}\) =2.5 (M)
a,PTHH: \(Ca\left(OH\right)_2+CO2\rightarrow CaCO_3+H_2O\)
b,\(n_{CO_2}=0,25\left(mol\right)\)
\(\Rightarrow n_{CaCO_3}=0,25mol\left(n_{CO_2}=n_{CaCO_3}\right)\)
\(\Rightarrow m_{CaCO_3}=0,25.100=25\)
c,\(C_{M_{Ca\left(OH\right)_2}}=2,5M\)
a)Ca(OH)2 + CO2--->CaCO3 +H2O
b) Ta có
n\(_{CO2}=\frac{5,6}{22,4}=0,25\left(mol\right)\)
Theo pthh
n\(_{CaCO3}=n_{CO2}=0,25\left(mol\right)\)
m\(_{CaCO3}=0,25.100=25\left(g\right)\)
c) Theo pthh
n\(_{Ca\left(OH2\right)}=n_{CO2}=0,25\left(mol\right)\)
C\(_M=\frac{0,25}{0,1}=2,5\left(M\right)\)
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