For the neutralization reaction involving
HNO3 and Ca(OH)2, how many liters of 1.55
M HNO3 are needed to react with 45.8 mL of
a 4.66 M Ca(OH)2 solution?
1. 1.32 L
2. 0.330 L
3. 0.137 L
4. 0.0343 L
5. 0.662 L
6. 0.275 L
chem help?
2017-01-16 12:06 am
回答 (2)
2017-01-16 12:36 am
Equation: 2HNO₃ + Ca(OH)₂ → Ca(NO₃)₂ + 2H₂O
OR: HNO₃ : Ca(OH)₂ = 2 : 1
No. of moles of Ca(OH)₂ = (4.66 mol/L) × (45.8/1000 L) = 0.2134 mol
No. of moles of HNO₃ = (0.2134 mol) × 2 = 0.4268 mol
Volume of HNO₃ = (0.4268 mol) / (1.55 mol/L) = 0.275 L
The answer: 6. 0.275 L
OR: HNO₃ : Ca(OH)₂ = 2 : 1
No. of moles of Ca(OH)₂ = (4.66 mol/L) × (45.8/1000 L) = 0.2134 mol
No. of moles of HNO₃ = (0.2134 mol) × 2 = 0.4268 mol
Volume of HNO₃ = (0.4268 mol) / (1.55 mol/L) = 0.275 L
The answer: 6. 0.275 L
2017-01-16 12:30 am
Where is your balanced equation? Why should be do everything for you?
收錄日期: 2021-04-18 16:03:10
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