2H2 + O2 → 2H2O
First, we will find the moles :
n H2 = m H2 / Mr H2
n H2 = 10 / 2
n H2 = 5 mole
n O2 = m O2 / Mr O2
n O2 = 5 / 16
n O2 = 0.3125 mole
n H2 / coef. H2 > n O2 / coef. O2
So, O2 is the limiting reactant
The mass of water produced :
n H2O = (coef. H2O / coef. O2) • n O2
n H2O = (2/1) • 0.3125
n H2O = 0.625 mole
m H2O = n H2O • Mr H2O
m H2O = 0.625 • 18
m H2O = 11.25 gr
Excesses reactant :
n H2 = 5 - 0.625 = 4.375 mole
m H2 = n H2 • Mr H2
m H2 = 4.375 • 2
m H2 = 8.75 gr
Answer:
a. Concave down
Linear increasing
b. Increases the reaction rate
c. The reaction approaches the saturation point of the enzyme
Explanation:
a. For the reaction with enzyme, the shape is concave down. The action of the enzyme on the preferred substrate is initially very rapid and decreases as the enzyme becomes saturated and the ratio of products to substrate increases to approach an equilibrium rate of reaction
For the reaction without enzyme, the shape is linear and increasing. Increase in the concentration of the substrate will increase the number of effective collisions that lead into product formation leading to an increased rate of the chemical reaction
b. The enzyme increases the proportion of effective combination of substrates to form the products
c. The curve of the reaction with enzyme flattens out because as the concentration of the substrate increases while that of the enzyme remains the same, the enzyme becomes saturated and less able to increase the rate of the reaction of the excess substrate.
C. Thermosphere is the correct answer
