How many mL of 0.471 M HNO3 are needed to dissolve 5.10 g of BaCO3?
回答 (1)
2016-05-30 10:31 pm
2HNO₃(aq) + BaCO₃(s) → Ba(NO₃)₂(aq) + H₂O(l) + CO₂(g)
Mole ratio HNO₃(aq) : BaCO₃ = 2 : 1
Molar mass of BaCO₃ = (137.3 + 12.01 + 16.00×3) g/mol = 197.3 g/mol
Number of moles of BaCO₃ = (5.10 g) / (197.3 g/mol) = 0.0258 mol
Number of moles of HNO₃ = (0.0258 mol) × 2 = 0.0516 mol
Volume of HNO₃ = (0.0516 mol) / (0.471 mol/L) = 0.110 L = 110 mL
Mole ratio HNO₃(aq) : BaCO₃ = 2 : 1
Molar mass of BaCO₃ = (137.3 + 12.01 + 16.00×3) g/mol = 197.3 g/mol
Number of moles of BaCO₃ = (5.10 g) / (197.3 g/mol) = 0.0258 mol
Number of moles of HNO₃ = (0.0258 mol) × 2 = 0.0516 mol
Volume of HNO₃ = (0.0516 mol) / (0.471 mol/L) = 0.110 L = 110 mL
收錄日期: 2021-04-18 14:57:50
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