nitrogen gas and oxygen gas combines to produce nitrogen dioxide. What amount of oxygen gas is needed to completely react with 3kg of nitrogen gas?
Here is that same stoichiometry problem example.Just follow the steps.
thanks foe helping me, but there are no steps shown ????
C'mon. I wrote the thing. I ought to know. I just looked (AGAIN) and I see
Step 1 in bold face type.
Step 2 in bold face type.
Step 3 in bold face type.
Step 4 in bold face type.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To determine the amount of oxygen gas needed to completely react with 3 kg of nitrogen gas, we need to use the balanced chemical equation for the reaction between nitrogen gas (N2) and oxygen gas (O2) to form nitrogen dioxide (NO2).
The balanced chemical equation for the reaction is:
2N2 + 4O2 → 4NO2
This equation shows that 2 moles of nitrogen gas react with 4 moles of oxygen gas to produce 4 moles of nitrogen dioxide.
To solve the problem, we can follow these steps:
1. Convert the mass of nitrogen gas (3 kg) to moles. To do this, we need to know the molar mass of nitrogen gas, which is 28.02 g/mol (14.01 g/mol for each nitrogen atom, multiplied by 2 since there are 2 nitrogen atoms in N2). Therefore, the number of moles of nitrogen gas is:
(3 kg) / (28.02 g/mol) = 107.1 mol
2. Since the balanced equation tells us that 2 moles of nitrogen gas react with 4 moles of oxygen gas, we can set up a proportion to find the number of moles of oxygen gas needed. The proportion can be set up as follows:
(2 mol N2) / (4 mol O2) = (107.1 mol N2) / (x mol O2)
Simplifying the proportion:
2 / 4 = 107.1 / x
Cross-multiplying:
2x = (4)(107.1)
x = (4)(107.1) / 2
x = 214.2 mol
Therefore, 214.2 moles of oxygen gas (O2) are needed to completely react with 3 kg of nitrogen gas (N2).