3g of H2 react with 29g of O2 to yield H2O.which is the limiting reactant? Calculate the maximum amount of H2O that can be formed. Calculate the amount the reactant which remains unreacted.

3 g of hydrogen will be the limiting reagent.
$2H_2\left(g\right)+O_2⟶2H_{2}O$
From the above equation, it is clear that 2 mole $H_2$ reacts with 1 mole $O_2$

The molar mass of $H_2$ = 2 g

The molar mass of $O_2$ = 32 g

​This implies,
4 g $H_2$ react with 32 g $O_2$
3 g $H_2$ reacts with = (32/4) x 3 g of $O_2$ gas
= 24 g

As the given amount of $O_2$ is more than required therefore $O_2$ is the excess reagent and $H_2$ is the limiting reagent.

2 mole of hydrogen gas reacts to form 2 mole of the water molecule, therefore,
4 g of $H_2$ produces = 36 g of water

So the amount of $H_{2}O$ produced by 3 g $H_2$ = (36/4) x 3
= 27 g
Hence, 27 g of water will be produced during the reaction
As 24 g of oxygen has been utilized during the reaction and 29 g of oxygen was supplied therefore the amount of oxygen gas left is (29-24) = 5g