Answer:
%Cl would be too high if not dried.
Explanation:
Cl (chlorine) is a liquid. During the process of drying AgCl, the chlorine particles evaporate and chlorine is lost.
Balanced Eqn
2
C
2
H
6
+
7
O
2
=
4
C
O
2
+
6
H
2
O
By the Balanced eqn
60g ethane requires 7x32= 224g oxygen
here ethane is in excess.oxygen will be fully consumed
hence
300g oxygen will consume
60
⋅
300
224
=
80.36
g
ethane
leaving (270-80.36)= 189.64 g ethane.
By the Balanced eqn
60g ethane produces 4x44 g CO2
hence amount of CO2 produced =
4
⋅
44
⋅
80.36
60
=
235.72
g
and its no. of moles will be
235.72
44
=5.36 where 44 is the molar mass of Carbon dioxide
hope this helps
Answer:
well, first off. the formula for carbon tetrachloride is CCl4
We need to find the molar mass of carbon tetrachloride
1(Mass of C) + 4(mass of chlorine)
1(12) + 4(35.5)
12 + 142
154 g/mol
Number of moles of CCl3 in 543.2g CCl3
n = given mass / molar mass
n = 543.2/153
n = 3.53 moles
always remember to brainly the questions you find helpful
Answer:
Mass = 8.46 g
Explanation:
Given data:
Mass of water produced = ?
Mass of glucose = 20 g
Mass of oxygen = 15 g
Solution:
Chemical equation:
C₆H₁₂O₆ + 6O₂ → 6H₂O + 6CO₂
Number of moles of glucose:
Number of moles = mass/molar mass
Number of moles = 20 g/ 180.16 g/mol
Number of moles = 0.11 mol
Number of moles of oxygen:
Number of moles = mass/molar mass
Number of moles = 15 g/ 32 g/mol
Number of moles = 0.47 mol
now we will compare the moles of water with oxygen and glucose.
C₆H₁₂O₆ : H₂O
1 : 6
0.11 : 6/1×0.11 = 0.66
O₂ : H₂O
6 : 6
0.47 : 0.47
Less number of moles of water are produced by oxygen thus it will limit the yield of water and act as limiting reactant.
Mass of water produced:
Mass = number of moles × molar mass
Mass = 0.47 mol ×18 g/mol
Mass = 8.46 g
V2 = 250 ml
Explanation:
Given:
P1 = 0.99 atm. V1 = 240 ml
P2 = 0.951 atm. V2 = ?
We can use Boyle's law to solve for V2
P1V1 = P2V2
V2 = (P1/P2)V1
= (0.99 atm/0.951 atm)(240 ml)
= 250. ml
