3g of hydrogen react with 29g of oxygen to produce water according to the given equation 2H2+O2=2H2O a)identify the limiting reagent b)calculate the mass of h2o formed

2 H ₂            + 0₂         ?  2 H₂O

2 X 2.01          32 g            2 x (2.01 + 16 ) = 36 .032

4.03 g

 

3 g of H ₂ needs  O_{2  =}  32 

                               --------   X 3  = 23 .8 g

                                 4.032

 

S0, 29 g 0f Oxygen is in excess, and hydrogen is the limiting reagent.

 

H₂O formed =  36.032  x 3    = 26.8 g

                   --------

                       4.032

 

 

O₂ left unreacted = 29 - 23.8

                         =  5.2 g