How many moles of NO(g) can be produced in the reaction of 3.00 mol NH3(g) and 4.00 mol O2(g) ?
4 NH3(g) + 5 O2(g) = 4 NO(g) + 6 H2O(l)
3 moles of NH"_3" requires "\\frac{3}{4}\u00d75=3.75\\>moles" of "O_2"
4 moles of O2 requires "\\frac{4}{5}\u00d74=3.2" moles of NH3
Limiting reagent is NH3
Moles of NO "=" moles of NH3
"= 3" moles
Comments
Leave a comment