What is needed to calculate the mass of ammonia gas produced from 2.0 L of nitrogen gas in excess hydrogen gas in the reaction below?
N2(g) + 3H2(g) ® 2NH3(g)?
Nothing but the coefficients in the balanced equation and convert to grams.
2.0 L N2 x (2 moles NH3/1 mole N2) = L NH3.
Convert L NH3 to moles (there are 22.4 L in a mole), then g = moles x molar mass.
To calculate the mass of ammonia gas produced from 2.0 L of nitrogen gas in excess hydrogen gas, you will need to use the following steps:
Step 1: Convert volume to moles (if necessary):
Since nitrogen gas (N2) and hydrogen gas (H2) are both gases, we need to convert the given volume of 2.0 L to moles. To do this, use the ideal gas equation:
PV = nRT
Where:
P = pressure (assume constant)
V = volume (2.0 L in this case)
n = number of moles
R = ideal gas constant
T = temperature (assume constant)
Step 2: Use stoichiometry to find the moles of ammonia (NH3) produced:
From the balanced chemical equation, we can see that the stoichiometric ratio between nitrogen gas (N2) and ammonia gas (NH3) is 1:2. This means that for every mole of nitrogen gas, 2 moles of ammonia gas are produced.
Therefore, if we find the moles of nitrogen gas using step 1, we can multiply it by the stoichiometric ratio to find the moles of ammonia gas.
Step 3: Convert moles to mass:
Now that we have the moles of ammonia gas, we can convert it to mass using the molar mass of ammonia gas (NH3). The molar mass of NH3 is calculated as the sum of the atomic masses of nitrogen (N) and hydrogen (H) multiplied by their respective subscripts in the balanced chemical equation.
Step 4: Perform the calculations:
Follow these steps to calculate the mass of ammonia gas produced:
1. Convert the volume of nitrogen gas to moles using the ideal gas equation.
2. Multiply the moles of nitrogen gas by the stoichiometric ratio (2 moles of NH3 per mole of N2) to obtain the moles of ammonia gas.
3. Convert the moles of ammonia gas to mass using the molar mass of NH3.
By following these steps, you will be able to calculate the mass of ammonia gas produced.
To calculate the mass of ammonia gas produced from 2.0 L of nitrogen gas in excess hydrogen gas, you will need the following information:
1. Balanced chemical equation: The given reaction is N2(g) + 3H2(g) -> 2NH3(g). This equation shows the stoichiometric ratio between nitrogen gas (N2), hydrogen gas (H2), and ammonia gas (NH3). It tells us that 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia.
2. Molar mass: You will need the molar mass of ammonia (NH3), nitrogen (N2), and hydrogen (H2). The molar mass is the mass of one mole of the substance and is typically expressed in grams/mole. The molar mass of ammonia (NH3) is approximately 17.03 g/mol, nitrogen (N2) is around 28.01 g/mol, and hydrogen (H2) is about 2.02 g/mol.
3. Volume of the gas: You have been provided with the volume of nitrogen gas, which is 2.0 L. This information will be used to calculate the number of moles of nitrogen.
4. Molar volume: The molar volume is the volume occupied by one mole of any gas at a specific temperature and pressure. At standard temperature and pressure (STP), the molar volume is approximately 22.4 L/mol. This is important because it allows us to convert between the volume of a gas and the number of moles.
Now, let's calculate the mass of ammonia gas produced:
1. Determine the number of moles of nitrogen gas (N2) using the ideal gas law. The ideal gas law equation is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. Since the problem does not provide pressure or temperature, we assume STP conditions.
At STP, 1 mole of gas occupies 22.4 L. Therefore, the number of moles of N2 gas can be calculated as:
(2.0 L) / (22.4 L/mol) = 0.0893 mol of N2
2. According to the balanced chemical equation, 1 mole of N2 reacts with 2 moles of NH3. Therefore, the number of moles of NH3 produced is:
(0.0893 mol N2) x (2 mol NH3 / 1 mol N2) = 0.1786 mol NH3
3. Finally, calculate the mass of ammonia gas produced (NH3) using its molar mass:
Mass = number of moles x molar mass
Mass of NH3 = 0.1786 mol NH3 x 17.03 g/mol NH3 = 3.04 g
Therefore, the mass of ammonia gas produced from 2.0 L of nitrogen gas in excess hydrogen gas is approximately 3.04 grams.