C3H8+3O2--->3CO2+8H
Therefore for every 1:3 there are 3 Carbon dioxides that form. That means find the limiting reactant from the two reactants.
5.5g(1mole C3H8/44.03g of C3H8)=0.1249 moled of C3H8 and if for every one C3H8 we can form three CO2. We can assume 0.3747 miles of CO2 will be produced.
15g of O2(1 mole O2/32g of O2)=0.4685moles O2 and if for every three O2 we can produce three CO2 we may assume a 1:1 ratio.
This means C3H8 will be your limiting reactant. Therefore 0.3747 moles of CO2 will be produced.
0.3747 moles of CO2(48.01 g of CO2/1 mole of CO2)= 17.99 grams of CO2
Answer: 2.58 days
Explanation:
Expression for rate law for first order kinetics is given by:
where,
k = rate constant = ?
t = age of sample = 6 days
a = initial amount of the reactant = 1 g
a - x = amount left after decay process
= 0.2 g
a) to find the rate constant
b) for completion of half life:
Half life is the amount of time taken by a radioactive material to decay to half of its original value.
The half life is 2.58 days
Answer:
the concentration of PCl5 in the equilibrium mixture = 296.20M
Explanation:
The concept of equilibrium constant was applied where the equilibrium constant is the ration of the concentration of the product over the concentration of the reactants raised to the power of their coefficients. it can be in terms of concentration in M or in terms of Pressure in atm.
The detaied steps is as shown in the attached file.
C. Weak acid is the correct answer
117.22 g are needed to react with an excess of Fe2O3 to produce 156.2 g of Fe.
Explanation:
Moles of Fe = Mass of Fe in grams / Atomic weight of Fe
= 156.2 / 55.847
Moles of Fe = 2.79.
The ratio between CO and Fe id 3 : 2.
Moles CO needed = 2.79 * (3 / 2)
= 4.185.
To calculate Atomic weight of CO,
Atomic weight of carbon = 12.011
Atomic weight of oxygen= 15.9994
Atomic weight of CO = 12.011 + 15.9994 = 28.01 g / mol.
Mass of CO = 4.185 * 28.01 = 117.22 g.