2Ca + O2 = 2CaO
First, determine which is the excess reactant
72.5 g Ca (1 mol) =1.8089725036
(40.078 g)
65 g O2 (1 mol) =2.0313769611
(15.999g × 2)
Since the ratio of to O2 is 2:1 in the balanced reaction, divide Ca's molar mass by 2 to get 0.9044862518. this isn't necessary because Ca is already obviously the limiting reactant. therefore, O2 is the excess reactant.
Now do the stoichiometry
72.5 g Ca (1 mol Ca) (1 mol O2)
(40.078 g Ca)(2 mol Ca)(31.998g O2)
=0.0282669621 g of O2 left over
Answer:
992.302 K
Explanation:
V(rms) = 750 m/s
V(rms) = √(3RT / M)
V = velocity of the gas
R = ideal gas constant = 8.314 J/mol.K
T = temperature of the gas
M = molar mass of the gas
Molar mass of CO₂ = [12 + (16*2)] = 12+32 = 44g/mol
Molar mass = 0.044kg/mol
From
½ M*V² = 3 / 2 RT
MV² = 3RT
K = constant
V² = 3RT / M
V = √(3RT / M)
So, from V = √(3RT / M)
V² = 3RT / M
V² * M = 3RT
T = (V² * M) / 3R
T = (750² * 0.044) / 3 * 8.314
T = 24750000 / 24.942
T = 992.302K
The temperature of the gas is 992.302K
Note : molar mass of the gas was converted from g/mol to kg/mol so the value can change depending on whichever one you use.
