Consider the equation N2(g)+3NH2(g)---2NH3.what is the needed volume of nitrogen to react with 15cm3 of hydrogen?
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I think the answer is 17.5 that is the neede volume
To determine the needed volume of nitrogen gas (N2) to react with 15 cm3 of hydrogen gas (H2), we need to first understand the balanced equation and use stoichiometry.
The balanced equation is: N2(g) + 3H2(g) ---> 2NH3(g)
From the equation, we can see that the ratio of N2 to H2 is 1:3. This means that for every 1 molecule of N2, we need 3 molecules of H2 to react, and vice versa.
Given that we have 15 cm3 of H2, we first need to find the amount of N2 needed to react with it.
Step 1: Convert the volume of H2 to the number of moles using the ideal gas equation.
The ideal gas equation is: PV = nRT
Where:
P = Pressure (in atm)
V = Volume (in L)
n = Number of moles
R = Gas constant (0.0821 L atm/(mol K))
T = Temperature (in Kelvin)
Assuming the gas pressure, temperature, and volume are constant, we can simplify the equation to:
n = (PV) / RT, where n is the number of moles
Since we're given the volume of hydrogen in cm3, we first need to convert it to liters.
15 cm3 = 15 / 1000 L = 0.015 L
Step 2: Calculate the number of moles of hydrogen gas (H2).
Assuming the pressure and temperature are constant and given that the molar volume of any gas at standard temperature and pressure (STP) is 22.4 L/mol, we can use the following equation:
n(H2) = V(H2) / 22.4
n(H2) = 0.015 L / 22.4 L/mol
Step 3: Determine the number of moles of nitrogen gas (N2) needed.
From the balanced equation, we know that for every 3 moles of H2, we need 1 mole of N2. Therefore, we can set up a proportion:
1 mole of N2 / 3 moles of H2 = n(N2) / n(H2)
Simplifying the equation gives us:
n(N2) = n(H2) * (1 mole of N2 / 3 moles of H2)
n(N2) = (0.015 L / 22.4 L/mol) * (1 mole of N2 / 3 moles of H2)
Step 4: Convert moles of N2 to the volume of N2.
To find the volume of N2, we can use the molar volume at STP (22.4 L/mol):
V(N2) = n(N2) * 22.4 L/mol
V(N2) = [(0.015 L / 22.4 L/mol) * (1 mole of N2 / 3 moles of H2)] * 22.4 L/mol
Simplifying the equation gives us:
V(N2) = (0.015 L * 1/3 * 22.4 L/mol) / 22.4 L/mol
V(N2) = 0.005 L
Therefore, the needed volume of nitrogen gas (N2) to react with 15 cm3 of hydrogen gas (H2) is 0.005 L.