N2+3H2>2NH3 that occurs at STP What volume of NH3 gas will form if 3g N2 react with enough H2?
To find the volume of NH3 gas that will form, we need to use the ideal gas law and the stoichiometry of the balanced chemical equation.
Step 1: Convert the mass of N2 to moles.
Given that the molar mass of N2 is approximately 28 g/mol:
moles of N2 = mass of N2 / molar mass of N2
moles of N2 = 3 g / 28 g/mol ≈ 0.1071 mol
Step 2: Determine the mole ratio between N2 and NH3.
From the balanced equation, we can see that 1 mole of N2 reacts to form 2 moles of NH3. Hence, the mole ratio is 1:2.
Step 3: Calculate the moles of NH3.
moles of NH3 = moles of N2 × (2 moles of NH3 / 1 mole of N2)
moles of NH3 = 0.1071 mol × (2 mol/1 mol) = 0.2142 mol
Step 4: Use the ideal gas law to find the volume of NH3 gas.
The ideal gas law equation is PV = nRT, where:
P is the pressure (which is STP, standardized to 1 atm),
V is the volume,
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 (0 °C = 273 K).
Substituting the given values:
P = 1 atm
n = 0.2142 mol
R = 0.0821 L·atm/(mol·K)
T = 273 K
Rearranging the equation to solve for V:
V = (n × R × T) / P
V = (0.2142 mol × 0.0821 L·atm/(mol·K) × 273 K) / 1 atm
V ≈ 4.7 L
Therefore, approximately 4.7 liters of NH3 gas will form when 3 grams of N2 reacts with enough H2 at STP.