consider-the-reaction-below-2-nh3-ocl-n2h4-cl-h2ocalculate-the-moles-of-n2h4-as-

Question

Consider the reaction below: 2 NH_{3} + $$OCl^{-}$$ → $$N_{2}H$$_{4} + $$Cl^{-}$$ + $$H_{2}$$OCalculate the moles of $$N_{2}$H_{4 }as $$produced if 6.93 mol of NH_{3} reacted and the reaction has a 85.6% yield

Accepted Answer

2.97 mol

Answer

4.56 mol

Answer

3.69 mol

Answer

8.96 mol

Consider the reaction below: 2 NH3 + OCl- → N2H4 + Cl- + H2OCalculate the moles of N2H4 as produced if 6.93 mol of NH3 reacted and the reaction has a 85.6% yield

Consider the reaction below:
2 NH₃ + OCl^- → N₂H₄ + Cl^- + H₂O
Calculate the moles of N₂H₄as produced if 6.93 mol of NH₃ reacted and the reaction has a 85.6% yield