The balanced reaction is
2CH3OH (g) + 3O2 (g) ⟶ 2CO2 (g) + 4H2O (g)
as per equation two moles of methanol gas will react with 3 moles of oxygen
one mole of gas occupies 22.4 L of volume
so the moles and volume goes in same ratio
it means two unit volume of methanol will react with three unit volume of oxygen
therefore 1L of methanol gas will react with 3 /2 L of oxygen
Or 18 L of methanol gas will react with 3 x 18 /2 = 27 L of oxygen
So here oxygen is limiting reagent
As per balanced equation
10L of oxygen will react with = 2 X 10 /3 L of methanol = 6.67 L of methanol gas to give 6.67 L of CO2 gas and 13.33 L of water gas
So overall there will be = 18 - 6.67 L of left out methanol = 11.33 L
And 6.67 L of CO2 + 13.33 L of water = 20 L
Total volume of gas = 11.33+ 20 = 31.33 L
Answer: Option (2) is the correct answer.
Explanation:
A real gas behaves least like an ideal gas under the conditions of low temperature and high pressure.
This is because at low temperature and high pressure molecules of gas will have negligible kinetic energy and strong force of attraction. Thus, real gas will not behave like an ideal gas.
Whereas at high temperature and low pressure a real gas will behave like an ideal gas.
I think the reaction involved here is a redox reaction in acidic conditions. To balance this type of reaction, we do as follows:
Balance the O atoms in each side by adding H2O on either side.
Balance H atoms in each side by adding H+.
Balance the charges by adding electrons.
Hope this helps.
The reaction between NaOH and HCl is as follows
NaOH + HCl ---> NaCl + H₂O
for neutralisation, H⁺ ions react with an equivalent amount of OH⁻ ions.
Number of NaOH moles reacted = 0.270 M/1000 mL/L x 37 mL = 0.00999 mol
number of HCl moles reacted = 0.270 M/1000 mL x 27 mL = 0.00729 mol
HCl reacts with NaOH in 1:1 molar ratio
Number of excess NaOH moles remaining - 0.00999 - 0.00729 = 0.0027 mol
total volume of solution = 37 mL + 27 mL = 64 mL = 0.064 L
Since there's excess OH⁻ ions, we can calculate pOH value first
pOH = - log [OH⁻]
[OH⁻] = 0.0027 mol / 0.064 L = 0.042 mol/L
pOH = -log(0.042 M)
pOH = 1.37
by knowing pOH we can calculate pH using the following equation;
pH + pOH = 14
pH = 14 - 1.37
pH = 12.63
