Use the law of constant composition to complete the following table summarizing the amounts of nitrogen and oxygen produced upon the decomposition of the sample of dinitrogen monoxide. ...
Use the law of constant composition to complete the following table summarizing the amounts of nitrogen and oxygen produced upon the decomposition of the sample of dinitrogen monoxide.
Mass N2O Mass N Mass O
Sample A 2.85 g 1.82 g 1.03 g
Sample C_____ _____ 1.95 g
Sample D_____ 1.91 g _____
I understand that N2O = N+O what I don't understand is how to determine the ratio so that when I only have the total Oxygen or Nitrogen to obtain the other totals based on the laws of constant composition
The chemical formula of the compound is N2O. When this gas decomposes nitrogen and oxygen are released.
2N2O --> 2N2 + O2
The molar mass of N2O is 14*2 + 16 = 28 + 16 = 44 g/mole, and the molar mass of oxygen and nitrogen is 32 and 14 respectively. Two moles of N2O decompose to give 2 moles of N2 and one mole of O2. Considering the mass of the elements individually, 1 mole of N2O gives 2 moles of N and 1 mole of O
From the values given for sample A, 2.85/44 = 0.064 moles of N2O decompose to give 1.82/14 = = 0.065*2 = 0.13 moles of N which tallies; 1.03 g of O is 0.064 mole of O which again tallies.
Similarly, for sample C, the mass of O generated is 1.95 g which is 0.121 mole. The corresponding values of the mass of N2O and mass of N are 0.121*44 = 5.36 g and 0.121*14 = 1.70 g respectively.
For sample D, the mass of N released is 1.91 g which is equivalent to 0.136 mole. The corresponding values for the mass of N2O and O are 6 g and 2.18 g respectively.