How many mL of 0.681 M HNO3 are needed to dissolve 8.18 g of CaCO3?
回答 (1)
2016-02-24 6:46 am
Molar mass of CaCO₃ = 40 + 12 + 16×3 = 100 g/mol
CaCO₃ + 2HNO₃ → Ca(NO₃)₂ + H₂O + CO₂
1 mole of CaCO₃ reacts with 2 moles of HNO₃.
No. of moles of CaCO₃ = 8.18/100 = 0.0818 mol
No. of moles of HNO₃ = 0.0818 × 2 = 0.1636 mol
Volume of HNO₃ = 0.1636/0.681 = 0.240 L = 240 mL
CaCO₃ + 2HNO₃ → Ca(NO₃)₂ + H₂O + CO₂
1 mole of CaCO₃ reacts with 2 moles of HNO₃.
No. of moles of CaCO₃ = 8.18/100 = 0.0818 mol
No. of moles of HNO₃ = 0.0818 × 2 = 0.1636 mol
Volume of HNO₃ = 0.1636/0.681 = 0.240 L = 240 mL
收錄日期: 2021-04-18 14:29:47
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