A 6.5 g of zinc powder reacts completely with access HCL and 1.5 L of H2 are obtained find max volume of H2
I assume you meant EXCESS HCl.
Volume of H2 at what conditions of T and P.
To find the maximum volume of H2 gas produced when 6.5 g of zinc powder reacts completely with excess HCl, we need to use the concept of stoichiometry and the ideal gas law.
Step 1: Determine the balanced chemical equation for the reaction between zinc (Zn) and hydrochloric acid (HCl). The balanced equation is:
Zn + 2HCl → ZnCl2 + H2
This equation shows that 1 mole of zinc reacts with 2 moles of hydrochloric acid to produce 1 mole of hydrogen gas.
Step 2: Calculate the number of moles of zinc (Zn) using its molar mass. The molar mass of zinc is 65.38 g/mol. Therefore:
Number of moles of Zn = Mass of Zn / Molar mass of Zn
= 6.5 g / 65.38 g/mol
≈ 0.0995 mol
Step 3: Determine the number of moles of hydrogen gas (H2) based on the stoichiometric ratio with zinc. Since the balanced equation shows that the ratio of Zn to H2 is 1:1, the number of moles of hydrogen gas is the same as the number of moles of zinc:
Number of moles of H2 = 0.0995 mol
Step 4: Use the ideal gas law to calculate the volume of hydrogen gas at standard temperature and pressure (STP: 0°C or 273.15 K and 1 atm). The ideal gas law equation is:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in L)
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)
At STP, the pressure is 1 atm and the temperature is 273.15 K.
Volume of H2 = (Number of moles of H2) × (R) × (T) / (P)
= (0.0995 mol) × (0.0821 L·atm/(mol·K)) × (273.15 K) / (1 atm)
≈ 2.17 L
Therefore, the maximum volume of H2 gas obtained when 6.5 g of zinc powder reacts with excess HCl is approximately 2.17 liters.