In the reaction represented by the equation 2Na + 2H2O > 2NaOH + H2, how many grams of hydrogen are produced if 120. g of Na and 80.0 g of H2O are available?

The answer to your question is  177.8 g of NaOH

Explanation:

Balanced chemical reaction

               2 Na  +  2 H₂O   ⇒  2 NaOH  +  H₂

Data

Grams of H₂ = ?

Mass of Na = 120 g

Mass of H₂O = 80

Process

1.- Calculate the limiting reactant

Molar mass of Na = 2 x 23 = 46 g

Molar mass of H₂O = 2 x 18 = 36 g

Theoretical proportion Na/H₂O = 46/36 = 1.28

Experimental proportion Na/H₂O = 120/80 = 1.5

As the experimental proportion increases, we conclude that the limiting reactant is the water.

2.- Calculate the mass of hydrogen

Molar mass of NaOH = 2[23 + 16 + 1] = 80 g

                     36 g of H₂O 80 g of NaOH

                      80 g of H₂O x

                      x = (80 x 80) / 36

                      x = 6400 / 36

                     x = 177.8 g of NaOH

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.