So so sorry if i'm wrong but i'm pretty sure no
It would be 0.341 because if you add 0.229 and 0.112 it will be 0.341
Mass of KCl= 1.08 g
<h3>Further explanation</h3>
Given
1 g of K₂CO₃
Required
Mass of KCl
Solution
Reaction
K₂CO₃ +2HCl ⇒ 2KCl +H₂O + CO₂
mol of K₂CO₃(MW=138 g/mol) :
= 1 g : 138 g/mol
= 0.00725
From the equation, mol ratio K₂CO₃ : KCl = 1 : 2, so mol KCl :
= 2/1 x mol K₂CO₃
= 2/1 x 0.00725
= 0.0145
Mass of KCl(MW=74.5 g/mol) :
= mol x MW
= 0.0145 x 74.5
= 1.08 g
It would be the same amount. So, 45 ml of NaOH is required to be added to the 45 ml of HCI to neutralize the acid fully. Here is a brief calculation:
Firstly, here is your formula: M(HCI) x V(HCI) = M(NaOh) x V(NaOH)
With the values put in: 0.35 x 45 = 0.35 x V(NaOH)
= 45 ml.
There is 45 ml of V(NaOH)
Let me know if you need anything else. :)
- Dotz
The volume of SO2 produced at 325k is calculated as below
calculate the moles of SO2 produced which is calculated as follows
write the reacting equation
K2SO3 +2 HCl = 2KCl +H2O+ SO2
find the moles of HCl used
=mass/molar mass = 15g/ 36.5 g/mol =0.411 moles
by use of mole ratio between HCl to SO2 which is 2:1 the moles of SO2 is therefore = 0.411 /2 =0.206 moles of SO2
use the idea gas equation to calculate the volume SO2
that is V=nRT/P
where n=0.206 moles
R(gas constant) = 0.082 L.atm/ mol.k
T=325 K
P=1.35 atm
V=(0.206 moles x 0.082 L.atm/mol.k x325 k)/1.35 atm = 4.07 L of SO2
