The following reaction is investigated (assume an Ideal gas mixture)2N2O(g) + N2H4(g) <--> 3N2(g) + 2H2O(g), Initially there are 0.10 mol of N2O and 0.25 mol of N2H4 in a 10.0-l container, If there are 0.50 mol of N20 at equalibrium, how many moles of N2 are present at equalibrium? please show all steps to answer this problem with significant figures