7.10 ml of 6.0 M HCL are added to 0.186g Zn(s) in a test tube. The balanced equa
ID: 792980 • Letter: 7
Question
7.10 ml of 6.0 M HCL are added to 0.186g Zn(s) in a test tube.The balanced equation for the reaction taking place is the following:
Zn+ 2 HCl ---> ZnCl2 + H2
The liberated H2 gas is collected by displacing water from an audiometer, a graduated glass measuring tube for gas analysis. After the reaction is complete, the volume of H2 collected above water is 72.18 ml at 18.8 7.10 ml of 6.0 M HCL are added to 0.186g Zn(s) in a test tube.
The balanced equation for the reaction taking place is the following:
Zn+ 2 HCl ---> ZnCl2 + H2
The liberated H2 gas is collected by displacing water from an audiometer, a graduated glass measuring tube for gas analysis. After the reaction is complete, the volume of H2 collected above water is 72.18 ml at 18.8
Explanation / Answer
convert ml to l
1000ml = 1l
10ml = 0.01l
convert M to mol
Molarity = moles of solute/liters of solution
6.0M = Xmoles/0.01l
Moles = 0.06
convert gram of zinc to moles
Moles = given mass/gram-fprmula
Moles = 0.186g/65gmol
Moles = 0.003
______________________________________...
1mol of zinc = 2 mols of HCl
0.06mol = (0.003)*2
0.06mol = 0.006
0.006Zn + 0.06HCl----->0.006ZnCl2 + 0.006H2
PV = nRT
n = 1mol
17.54 * 72.18 = 1*R * 18.8
1266 = 18.8R
R = 67.3
Related Questions
Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.