A volume of a pure sample of HCl gas was 289mL at 25 C and 208 mmHg. The HCl gas was completely dissolved in 70.0mL of water and titrated with NaOH solution. What would be the resulting molarity of HCl?
If 19.43mL of the NaOH solution were required to neutralize the HCl, what is the molarity of the NaOH solution?
Then another set;
Nitrogen monoxide reacts with molecular oxygen as follows: 2NO (g) + O2 (g) -> 2NO2(g). Initially NO and O2 are separated as shown in the diagram labeled FIGURE B (NO is 3.00L and 0.465atm and O2 is 1.75L and 1.00atm). When the valve is opened, the reaction quickly goes to completion. Assume the gases remain at 25 C before and after the reaction.
Determine what gases remain at the end of the reaction.
Calculate the moles of gases that remain at the end of the reaction.
Calculate the partial pressures of all the gases that remain after the reaction is complete.
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2017-12-31 9:12 am
回答 (2)
2017-12-31 10:34 am
✔ 最佳答案
Consider the HCl gas :Pressure, P = 208/760 atm
Volume, V = 289 mL = 0.289 L
Gas constant, R = 0.08206 L atm / (mol K)
Temperature, T = (273 + 25) K = 298 K
Gas law : PV = nRT
No. of moles of HCl, n = PV/(RT) = (208/760) × 0.289 / (0.08206 × 298) mol = 0.003234 mol
Resulting molarity of HCl = (0.003234 mol) / (70/1000 L) = 0.0462 M
Equation for the reaction between HCl and NaOH :
HCl + NaOH → NaCl + H₂O
Molar ratio HCl : NaOH = 1 : 1
No. of moles of HCl reacted = 0.003234 mol
No. of millimoles of NaOH required = 0.003234 mol
Molarity of NaOH = (0.003234 mol) / (19.43/1000 L) = 0.166 M
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When the valve is open and before the reaction occurs :
Initial partial pressure of NO = (0.465 atm) × [3.00/(3.00 + 1.75)] = 0.2937 atm
Initial partial pressure of O₂ = (1.00 atm) × [1.75/(300 + 1.75)] = 0.3684 atm
Equation for the reaction :
2NO + O₂ → 2NO₂
Mole ratio NO : O₂ = 2 : 1 : 2
More NO is needed, but (Initial partial pressure of NO) < (Initial partial pressure of O₂)
Hence, NO is the limiting reactant/reagent, and O₂ is in excess.
Partial pressure of NO after reaction = 0 atm
Partial pressure of O₂ after reaction = [0.3684 - 0.2937 × (1/2)] atm = 0.2216 atm
Partial pressure of NO₂ after reaction = 0.2937 × (2/2) atm = 0.2937 atm
The gaseous mixture after reaction :
Pressure, P = (0.2216 + 0.2937) atm = 0.5153 atm
Volume, V = (3.00 + 1.75) L = 4.75 L
Gas constant, R = 0.08206 L atm / (mol K)
Temperature, T = (273 + 25) K = 298 K
Gas law : PV = nRT
No. of moles of gas after reaction, n = PV/(RT) = 0.5153 × 4.75 / (0.08206 × 298) mol = 0.100 mol
2017-12-31 10:17 am
A volume of a pure sample of HCl gas was 289mL at 25 C and 208 mmHg. The HCl gas was completely dissolved in 70.0mL of water and titrated with NaOH solution. What would be the resulting molarity of HCl?
PV = nRT ... n = [(208/760)(0.289)] / [(0.0821)(298)] = 0.00320 mol
M = moles/Liters = 0.00320/0.070 = 0.457 M <<< answer
If 19.43mL of the NaOH solution were required to neutralize the HCl, what is the molarity of the NaOH solution?
0.0032 = (0.01943)M
M = 0.1647M <<< molar conc. of NaOH <<< 2nd answer
These 2 were related so I did both. My general rules is ONE answer per post, so you try the other.
PV = nRT ... n = [(208/760)(0.289)] / [(0.0821)(298)] = 0.00320 mol
M = moles/Liters = 0.00320/0.070 = 0.457 M <<< answer
If 19.43mL of the NaOH solution were required to neutralize the HCl, what is the molarity of the NaOH solution?
0.0032 = (0.01943)M
M = 0.1647M <<< molar conc. of NaOH <<< 2nd answer
These 2 were related so I did both. My general rules is ONE answer per post, so you try the other.
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