1 N204(l) + 2 N2H4(l) --> 3 N2(g) + 4 H2O(g)
If you start with: 5.0 x 10^4 hydrazine (N2H4) "in the tank":
1) How many moles of nitrogen can be created (assuming 100% yield)?
2) How many moles of water can be produced?
2) What mass of dinitrogen tetroxide (N2O4) is needed? What volume of N2O4 is needed? (density = 1.44 g/cm^3)
Can anyone help me on these problems?
what is the 5.0x1064 hydrazine in? grams, moles, etc.
sorrry. in grams.
See my response to your later post.
1 N204(l) + 2 N2H4(l) --> 3 N2(g) + 4 H2O(g)
Certainly! Let's break down each question step by step.
1) To find the number of moles of nitrogen that can be created, we need to use the stoichiometry of the balanced chemical equation. From the equation, we can see that the molar ratio between N204 and N2 is 1:3.
Given that we started with 5.0 x 10^4 moles of N2H4, let's find how many moles of N2 can be produced. Since the ratio of N2H4 to N2 is 2:3, we multiply the number of moles of N2H4 by the ratio:
Number of moles of N2 = (5.0 x 10^4 moles N2H4) * (3 moles N2/2 moles N2H4)
Number of moles of N2 = 7.5 x 10^4 moles N2
Therefore, 7.5 x 10^4 moles of nitrogen can be created assuming 100% yield.
2) Similar to the previous question, we need to use the stoichiometry of the balanced chemical equation to determine the number of moles of water produced. From the equation, it can be seen that the molar ratio between N2H4 and H2O is 2:4 (or simplified, 1:2).
Number of moles of water = (5.0 x 10^4 moles N2H4) * (2 moles H2O/2 moles N2H4)
Number of moles of water = 5.0 x 10^4 moles H2O
Therefore, 5.0 x 10^4 moles of water can be produced assuming 100% yield.
3) To determine the mass and volume of dinitrogen tetroxide (N2O4) needed, we need to use its density (given as 1.44 g/cm^3). However, the given balanced equation does not mention N2O4. We need additional information or calculations to answer this question. Could you provide any additional information or equations related to the formation of N2O4 from the given reactants?