mCaCO3​,caˆˋn thieˆˊt​=MCaO​×HmCaO, thực teˆˊ​×MCaCO3​​​

Thay số vào công thức:
\(m_{\text{CaCO}_{3} , \text{c} \overset{ˋ}{\hat{\text{a}}} \text{n}\&\text{nbsp};\text{thi} \overset{ˊ}{\hat{\text{e}}} \text{t}} = \frac{7 \times 100}{56 \times 0.9} = \frac{700}{50.4}\)
\(m_{\text{CaCO}_{3} , \text{c} \overset{ˋ}{\hat{\text{a}}} \text{n}\&\text{nbsp};\text{thi} \overset{ˊ}{\hat{\text{e}}} \text{t}} \approx 13.888... \&\text{nbsp};\text{t} \overset{ˊ}{\hat{\text{a}}} \text{n}\)
\(m_{\text{qu}ặ\text{ng}} = \frac{m_{\text{CaCO}_{3} , \text{c} \overset{ˋ}{\hat{\text{a}}} \text{n}\&\text{nbsp};\text{thi} \overset{ˊ}{\hat{\text{e}}} \text{t}}}{\text{H} \overset{ˋ}{\text{a}} \text{m}\&\text{nbsp};\text{l}ượ\text{ng}\&\text{nbsp}; \text{CaCO}_{3}}\)
\(m_{\text{qu}ặ\text{ng}} = \frac{13.888...}{80 \%} = \frac{13.888...}{0.8}\)
\(m_{\text{qu}ặ\text{ng}} \approx 17.3611... \&\text{nbsp};\text{t} \overset{ˊ}{\hat{\text{a}}} \text{n}\)
\(m_{\text{qu}ặ\text{ng}} \approx 17.36 \&\text{nbsp};\text{t} \overset{ˊ}{\hat{\text{a}}} \text{n}\)

1. \(\text{Cu}(\text{OH})_{2}\): Hidroxit đồng
2. \(\text{Cu}(\text{OH})_{2}\): Hidroxit đồng
3. \(\text{BaSO}_{4}\): Sunfat bari
4. \(\text{BaSO}_{4}\): Sunfat bari

Cu+2AgNO3​→Cu(NO3​)2​+2Ag↓

2Na3​PO4​+3CaCl2​→Ca3​(PO4​)2​↓+6NaCl