Respuesta :
Answer:
For 1: The volume of carbon monoxide required to produce given amount of methanol is 11.9 L.
For 2: The volume of oxygen required to form given amount of water is 7.77 L.
Explanation:
- For 1:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)
For methanol:
Given mass of methanol = 19.3 g
Molar mass of methanol = 32 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of methanol}=\frac{19.3g}{32g/mol}=0.603mol[/tex]
The chemical reaction of formation of methanol from carbon monoxide follows:
[tex]CO+2H_2\rightarrow CH_3OH[/tex]
By Stoichiometry of the reaction:
1 mole of methanol is produced from 1 mole of carbon monoxide.
so, 0.603 moles of methanol will be produced from = [tex]\frac{1}{1}\times 0.603=0.603mol[/tex] of carbon monoxide.
To calculate the volume of carbon monoxide, we use the equation given by ideal gas:
PV = nRT
where,
P = Pressure of carbon monoxide = 676 mmHg
V = Volume of carbon monoxide = ? L
n = Number of moles of carbon monoxide = 0.603 mol
R = Gas constant = [tex]62.3637\text{ L.mmHg }mol^{-1}K^{-1}[/tex]
T = Temperature of carbon monoxide = 212 K
Putting values in above equation, we get:
[tex]676mmHg\times V=\frac{19.3}{32g/mol}\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 212K\\\\V=11.79L[/tex]
Hence, the volume of carbon monoxide required to produce given amount of methanol is 11.9 L.
- For 2:
Calculating the moles of water by using equation 1, we get:
Given mass of water = 12.5 g
Molar mass of water = 18 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of water}=\frac{12.5g}{18g/mol}=0.694mol[/tex]
The chemical reaction of formation of water from oxygen and hydrogen follows:
[tex]O_2+2H_2\rightarrow 2H_2O[/tex]
By Stoichiometry of the reaction:
2 moles of water is produced from 1 mole of oxygen gas.
so, 0.694 moles of water will be produced from = [tex]\frac{1}{2}\times 0.694=0.347mol[/tex] of oxygen gas.
At STP:
1 mole of a gas occupies 22.4 L of volume.
So, 0.347 moles of oxygen gas will occupy = [tex]\frac{22.4L}{1mol}\times 0.347mol=7.77L[/tex]
Hence, the volume of oxygen required to form given amount of water is 7.77 L.
