First we determine the
moles CaCl2 present:
525g / (110.9g/mole) =
4.73 moles CaCl2 present
Based on stoichiometry,
there are 2 moles of Cl for every mole of CaCl2:<span>
(2moles Cl / 1mole CaCl2) x 4.73 moles CaCl2 = 9.47 moles Cl </span>
Get the mass:<span>
<span>9.47moles Cl x 35.45g/mole = 335.64 g Cl</span></span>
Answer:
The answer to your question is:
1.- CO
2.- 0.414 moles of CO2
Explanation:
Data
2CO + O2 ⇒ 2CO2
CO = 0.414 moles
O2 = 0.418
Process
theoretical ratio CO/O2 = 2/1 = 1
experimental ratio CO/O2 = 0.414/0.418 = 0.99
Then the limiting reactant is CO
2.-
2 moles of CO --------------- 2 moles of CO2
0.414 moles of CO --------- x
x = (0.414 x 2) / 2
x = 0.414 moles of CO2
Q=m(c∆t +heat of fusion + heat of evaporation)
m= 44g
c= 4.186 J/g.C
∆t= 107-(-8) =115 C
heat of fusion= 333.55 J/g
heat of evaporation=2260 J/g
Q=44(4.186*115 + 333.55 + 2260)
Q= 135297.36 J
Answer:
The appropriate answer is "9.225 g".
Explanation:
Given:
Required level,
= 63 ppm
Initial concentration,
= 22 ppm
Now,
The amount of free SO₂ will be:
= 
= 
= 
The amount of free SO₂ to be added will be:
= 
= 
∵ 1000 mg = 1 g
So,
= 
= 
Thus,
"9.225 g" should be added.
