using this equation: Ni+H2SO4=H2+NiSO4
how many liters of hydrogen gas at 298degK and 2.5atm could be produced if you react 25 grams of nickel metal?
Follow the steps in this example, then use PV = nRT to calculate volume.
http://www.jiskha.com/science/chemistry/stoichiometry.html
I ended up getting 4.21L of H2 is that right?
The steps I took were
25gNi/58.6934=.43mol Ni
v=(.43)x(0.0821x1atm/molK)x(298K)/2.5atm=4.21L H2?
this doesn't look right...
To find out how many liters of hydrogen gas can be produced, we need to use stoichiometry, which relates the amount of reactants and products in a chemical equation.
First, let's calculate the number of moles of nickel (Ni) using its molar mass. The molar mass of nickel is 58.69 g/mol.
Number of moles of Ni = Given mass / Molar mass
Number of moles of Ni = 25 g / 58.69 g/mol
Next, we need to determine the mole ratio between Ni and H2 from the balanced chemical equation.
From the equation: Ni + H2SO4 → H2 + NiSO4
The mole ratio of Ni to H2 is 1:1. This means for every 1 mole of Ni, 1 mole of H2 is produced.
Now we can calculate the number of moles of H2:
Number of moles of H2 = Number of moles of Ni
Since we have the number of moles of H2, we can now use the ideal gas law to find the volume of H2 at 298 K and 2.5 atm.
Ideal Gas Law: PV = nRT
P is the pressure in atm, V is the volume in liters, n is the number of moles, R is the ideal gas constant (0.0821 L⋅atm/(mol⋅K)), and T is the temperature in Kelvin.
Number of liters of H2 = (Number of moles of H2 * R * T) / P
Plugging in the values:
Number of liters of H2 = (Number of moles of H2 * 0.0821 L⋅atm/(mol⋅K) * 298 K) / 2.5 atm
Now, let's calculate the final answer:
Number of liters of H2 = (25 g / 58.69 g/mol * 0.0821 L⋅atm/(mol⋅K) * 298 K) / 2.5 atm
After performing the calculations, the resulting value will be the number of liters of hydrogen gas that can be produced when 25 grams of nickel metal reacts.