C3H8 + 5O2 ------> 3CO2 + 4H2O
from reaction 1 mol 5 mol
given 1.82 mol x mol
x=(1.82*5)/1 = 9.10 mol CO2
The mass of titanium is = 47,867 g/1mol
Applying the rule of avrogado
1mol _______ 6,023 × 10^(23) at
0,075mol ___ x
X . 1mol = 0,075mol . 6,023 . 10^(23)at
X = 0,075 . 6,023 . 10^(23) at
X = 4,51 . 10^(22) atoms
Hope this helps
Answer:
:) ..................................................................:)
Volume of tank = 
(given)
Since, 
So,
For :
The significant rule for multiplication, states that the number of significant figures in the answer obtained by multiplication is determined by the value with the lowest number of significant digits.
Since, the minimum number of decimal places in the above multiplication operation is 1 so, the final result must be upto 1 decimal place only.
Hence, volume in 
is 243.5.
<span>atomic weights: Al = 26.98, Cl = 35.45
In this reaction; 2Al = 53.96 and 3Cl2 = 212.7
Ratio of Al:Cl = 53.96/212.7 = 0.2537 that is approximately four times the mass Cl is needed.
Step 2:
(a) Ratio of Al:Cl = 2.70/4.05 = 0.6667
since the ratio is greater than 0.2537 the divisor which is Cl is not big enough to give a smaller ratio equal to 0.2537.
so Cl is limiting
(b)since Cl is the limiting reactant 4.05g will be used to determine the mass of AlCl3 that can be produced.
From Step 1:
212.7g of Cl will produce 266.66g AlCl3
212.7g = 266.66g
4.05g = x
x = 5.08g of AlCl3 can be produced
(c)
Al:Cl = 0.2537
Al:Cl = Al:4.05 = 0.2537
mass of Al used in reaction = 4.05 x 0.2537 = 1.027g
Excess reactant = 2.70 - 1.027 = 1.67g
King Leo · 9 years ago</span>
