Question
Determine the volume of hydrogen gas formed when excess zinc metal is added to 100cm³ of one molar hydrochloric acid(1 mol of gas occupied 24.0 litres at room temp and pressure)
Answers
GPT 3.5
The balanced chemical equation for the reaction between zinc and hydrochloric acid is:
Zn + 2HCl -> ZnCl2 + H2
From the equation, we can see that for every mole of zinc used, one mole of hydrogen gas is produced.
Given that the volume of hydrochloric acid is 100 cm³ (which is equal to 0.1 liters) and its molarity is 1M, we can calculate the number of moles of hydrochloric acid:
Moles of HCl = volume (in liters) x molarity
= 0.1 L x 1 mol/L
= 0.1 mol
Since the reaction between zinc and hydrochloric acid is in a 1:1 ratio, the number of moles of hydrogen gas produced is also 0.1 mol.
By using the volume-mole relationship given (1 mol of gas occupies 24 liters), we can calculate the volume of hydrogen gas generated:
Volume of hydrogen gas = number of moles x 24 L/mol
= 0.1 mol x 24 L/mol
= 2.4 L
Therefore, the volume of hydrogen gas formed when excess zinc metal is added to 100 cm³ of 1M hydrochloric acid is 2.4 liters.
Zn + 2HCl -> ZnCl2 + H2
From the equation, we can see that for every mole of zinc used, one mole of hydrogen gas is produced.
Given that the volume of hydrochloric acid is 100 cm³ (which is equal to 0.1 liters) and its molarity is 1M, we can calculate the number of moles of hydrochloric acid:
Moles of HCl = volume (in liters) x molarity
= 0.1 L x 1 mol/L
= 0.1 mol
Since the reaction between zinc and hydrochloric acid is in a 1:1 ratio, the number of moles of hydrogen gas produced is also 0.1 mol.
By using the volume-mole relationship given (1 mol of gas occupies 24 liters), we can calculate the volume of hydrogen gas generated:
Volume of hydrogen gas = number of moles x 24 L/mol
= 0.1 mol x 24 L/mol
= 2.4 L
Therefore, the volume of hydrogen gas formed when excess zinc metal is added to 100 cm³ of 1M hydrochloric acid is 2.4 liters.