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:
Option B. 4 moles of the gaseous product
Explanation:
Data obtained from the question include:
Initial volume (V1) = V
Initial number of mole (n1) = 2 moles
Final volume (V2) = 2V
Final number of mole (n2) =..?
Applying the Avogadro's law equation, we can obtain the number of mole of the gaseous product as follow:
V1/n1 = V2/n2
V/2 = 2V/n2
Cross multiply
V x n2 = 2 x 2V
Divide both side by V
n2 = (2 x 2V)/V
n2 = 2 x 2
n2 = 4 moles
Therefore, 4 moles of the gaseous product were produced.
Answer:
1.3 meters
Explanation: use newton third law equation.
Answer:
135g Na2CO3
Explanation:
I'm going to assume you mean Molality which is mol solute/kg solvent
Molarity would be mol soute/ L solution
we know we have 155g of water which is .155 kg
essentially we have the equation:
mol/kg = 8.20
we substitute .155 in for kg and get:
mol/.155 = 8.20
Solving this gives mol = 1.271
now we must convert to grams using the molar mass
Molar mass Na2CO3 = 106G/mol
so to cancel moles we multiply:
1.271mol x 106g/mol
= 135g
