Question

Answer 1

Ta có: \(n_{CO_3^{2-}}=0,04+0,07=0,11\left(mol\right)\)

\(n_{Ca^{2+}}=0,02+0,08=0,1\left(mol\right)\)

PT: \(CO_3^{2-}+Ca^{2+}\rightarrow CaCO_3\)

____0,11___0,1 _______0,1 (mol)

\(\Rightarrow m_{CaCO_3\downarrow}=0,1.100=10\left(g\right)\)

Bạn tham khảo nhé!

Answer 2

\(n_{CO_3^{2-}\left(bđ\right)}=0,04+0,07=0,11\left(mol\right)\)

\(n_{Ca^{2+}}=0,02+0,08=0,1\left(mol\right)\)

\(Ca^{2+}+CO_3^{2-}\rightarrow CaCO_3\)

0,1.............0,11

Lập tỉ lệ \(\dfrac{0,1}{1}< \dfrac{0,11}{1}\) => Sau phản ứng CO3^2- dư

\(n_{CaCO_3}=n_{Ca^{2+}}=0,1\left(mol\right)\)

=> \(m_{CaCO_3}=0,1.100=10\left(g\right)\)