Respuesta :
Answer:
The pressure is assessed through the ideal gas equation and Dalton's law
Explanation:
Partial Pressure of hydrogen gas in the gas collecting tube can be calculated as:
According to Dalton's law, the total pressure of the system will be the sum of partial pressure of all gases.
In the system, the pressure is exerted by the water and [tex]\rm H_2[/tex] gas.
Total pressure = [tex]\RM P_water + P_H_2[/tex]
746 mm Hg (given) = 19.8 mm Hg + [tex]\rm P_H_2[/tex]
[tex]\rm P_H_2[/tex] = 746 mm Hg - 19.8 mm Hg
[tex]\rm P_H_2[/tex] = 726.2 mm Hg
So, the partial pressure of Hydrogen was found to be 726.2 mm Hg.
To calculate the volume of hydrogen at STP, ideal gas equation is used.
For this, we calculate the moles of Hydrogen in given conditions:
Pressure = 726.2 mm Hg
= [tex]\frac{726.2}{760}[/tex]
= 0.955 atm
Temperature = [tex]\rm 22^0\;C[/tex]
= 22 + 273.15 K
= 295.15 K
Volume = 31.0 ml
= [tex]\rm 31.0 \times 10^-^3[/tex] liters
Substituting the values in the ideal gas equation:
PV= nRT
[tex]\RM 0.955\; \times\; 31\;\times 10^-^3 =\; n\; 0.0821\; \times\; 295.15[/tex]
n = [tex]\RM 1.222\; \times\;10^-^3[/tex] moles
The volume of hydrogen for [tex]\RM 1.222\; \times\;10^-^3[/tex] moles of hydrogen at STP will be:
at STP
P = 1 atm
T = 273 K
[tex]\rm 1\; \times\; V\;=\;1.222\;\times\;10^-^3\;\times\;0.0821\;\times\;273[/tex]
V = 0.027398 liters
V = [tex]\rm 0.027398\; \times\;1000[/tex] ml
V = 27.398 ml
For producing [tex]\rm H_2[/tex] from 0.028 gram of Mg,
[tex]\rm Mg\;+\;2\;HCl\;\rightarrow\;MgCL_2\;+H_2[/tex]
1 mole of magnesium produces 1 mole of Hydrogen
Molar mass of Mg is 24.3 grams/ moles
Moles of Mg = [tex]\rm\frac{weight}{molecular weight}[/tex]
= [tex]\rm\frac{0.028}{24.3}[/tex] moles
= 0.001523 moles of Magnesium is involved in the reaction. Hence equal volume of [tex]\rm H_2[/tex] i.e. 0.001523 moles of hydrogen is produced.
For more information, refer the link:
https://brainly.com/question/8408970?referrer=searchResults
