what volume of ammonia gas will be produced when 1.26L of nitrogen react completely in the following equation?

N2 + 3H2 > 2NH3

Please if someone can just give me the order of operation it will be much appreciated because chemistry is just not my forte.

Just use the coefficients to tell you.

1.26 L N2 x (2 mol NH3/1 mol N2) = 1.26 x (2/1) = ?L NH3

From the equation,1L of N2 produced 2L of NH3.1.26L of N2 will produce 1.26X2/1=2.52L of NH3(Ammonia).

To find the volume of ammonia gas produced, we need to use the stoichiometry of the reaction and the given volume of nitrogen.

First, let's determine the stoichiometric ratio between nitrogen (N2) and ammonia (NH3) using the balanced equation:
N2 + 3H2 → 2NH3

From the balanced equation, we can see that for every 1 mole of nitrogen (N2), 2 moles of ammonia (NH3) are produced. This gives us a stoichiometric ratio of 1:2.

Next, we'll need to convert the given volume of nitrogen gas (1.26 L) into moles. To do this, we can use the ideal gas law equation:

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)

Since we are not given the pressure and temperature, we can assume standard temperature and pressure (STP), which is 1 atm and 273.15K. Therefore, the ideal gas law equation simplifies to:

V = n × 22.4 L/mol (at STP)

Now, let's calculate the number of moles of nitrogen gas:
n(N2) = V(N2)/22.4 L/mol

n(N2) = 1.26 L / 22.4 L/mol

Now, we can use the stoichiometric ratio to find the number of moles of ammonia gas produced:
n(NH3) = n(N2) × 2

Finally, we can convert the moles of ammonia gas back into volume:
V(NH3) = n(NH3) × 22.4 L/mol

Therefore, the volume of ammonia gas produced can be calculated by multiplying the number of moles of ammonia by 22.4 L/mol.