Answer:
[tex]m_{N_2}=93.3gN_2[/tex]
Explanation:
Hello there!
In this case, for this stoichiometry-based problem, it is firstly necessary to realize that the decomposition of ammonium dichromate is given by:
[tex](NH_4)_2Cr_2O_7(s)\rightarrow N_2(g)+4H_2O(l)+Cr_2O_3(s)[/tex]
Thus, since the mole ratio between ammonium dichromate and the gaseous nitrogen (molar mass = 28.02 g/mol) is 1:1, we can compute the produced mass of the latter via stoichiometry as shown below:
[tex]m_{N_2}=3.33mol(NH_4)_2Cr_2O_7*\frac{1molN_2}{1mol(NH_4)_2Cr_2O_7}*\frac{28.01gN_2}{1molN_2}\\\\ m_{N_2}=93.3gN_2[/tex]
Best regards!
