a)
Có: \(m_{tăng}=m_{CO_2}=m_{\downarrow}\)
\(\rightarrow m_{CO_2}=15+2,6=17,6\left(g\right)\\ \rightarrow n_{CO_2}=\dfrac{17,6}{44}=0,4\left(mol\right)\)
=> V = 0,4.22,4 = 8,96 (l)
b)
\(n_{CaCO_3}=\dfrac{15}{100}=0,15\left(mol\right)\)
\(\xrightarrow[\text{BTNT C}]{}n_{CO_2\left(dư\right)}=0,4-0,15=0,25\left(mol\right)\)
PTHH:
Ca(OH)2 + CO2 ---> CaCO3 + H2O
0,15<--------------------0,15
Ca(OH)2 + 2CO2 ---> Ca(HCO3)2
0,125<----0,25
\(\rightarrow C_{M\left(Ca\left(OH\right)_2\right)}=\dfrac{0,125+0,15}{0,2}=1,375M\)
Đề bài phải cho thể tích CO2 ở điều kiện tiêu chuẩn thì mới tính được V nhé !
a) \(m_{\text{bình tăng}}=m_{CO_2}-m_{\text{kết tủa}}=m_{CO_2}-15=2,6\left(g\right)\\ \Rightarrow m_{CO_2}=17,6\left(g\right)\\ n_{CO_2}=\dfrac{17,6}{44}=0,4\left(mol\right)\\ V=V_{CO_2\left(ĐKTC\right)}=0,4\cdot22,4=8,96\left(l\right)\)
b) \(n_{CaCO_3}=\dfrac{15}{100}=0,15\left(mol\right)\)
\(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\left(1\right)\\ 2CO_2+Ca\left(OH\right)_2\rightarrow Ca\left(HCO_3\right)_2\left(2\right)\)
\(\left(1\right)\Rightarrow n_{CO_2\left(1\right)}=n_{Ca\left(OH\right)_2\left(1\right)}=n_{CaCO_3}=0,15\left(mol\right)\\ \Rightarrow n_{CO_2\left(2\right)}=0,4-0,15=0.25\left(mol\right)\\ \left(2\right)\Rightarrow n_{Ca\left(OH\right)_2\left(2\right)}=\dfrac{1}{2}n_{CO_2\left(2\right)}=\dfrac{1}{2}\cdot0,25=0,125\left(mol\right)\\ \Rightarrow n_{Ca\left(OH\right)_2}=0,15+0,125=0,275\left(mol\right)\\ C_{M_{Ca\left(OH\right)_2}}=\dfrac{0,275}{0,2}=1,375\left(M\right)\)
