For the following reaction, 76.0 grams of barium chloride are allowed to react with 67.0 grams of potassium sulfate. barium chloride (aq) potassium sulfate (aq) barium sulfate (s) potassium chloride (aq)

For the following reaction, 76.0 grams of barium chloride are allowed to react with 67.0 grams of potassium sulfate.

The reaction consumes _____ moles of barium chloride. The reaction produces _____ moles of barium sulfate and _____ moles of potassium chloride.

a) The reaction consumes 0.365 moles of barium chloride.

b) The reaction produces 0.365 moles of barium sulfate and 0.730 moles of potassium chloride.

Explanation:

To calculate the moles :

According to stoichiometry :

1 mole of require 1 mole of

Thus 0.365 moles of will require=  of

Thus is the limiting reagent as it limits the formation of product and is the excess reagent.

As 1 moles of give = 1 moles of

Thus 0.365 moles of give =  of

As 1 moles of give = 2 moles of

Thus 0.365 moles of give =  of

Thus the reaction consumes 0.365 moles of barium chloride. The reaction produces 0.365 moles of barium sulfate and 0.730 moles of potassium chloride.

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.