Answer :
Answer:-
182 Kg
Explanation:-
Molar mass of CO = 12 x 1 + 16 x 1 = 28 grams
Molar mass of CH3OH = 12 x 1 + 1 x 3 + 16 x 1 + 1 x 1
= 32 grams
The balanced equation for this reaction is
CO(g)+H2(g)—> CH3OH(g)
From the equation we see
1 CH3OH is produced from 1 CO
32 grams of CH3OH is produced from 28 grams of CO.
208 Kg of CH3OH is produced from (28 gram x 208 Kg) / (32 gram)
= 182 Kg of CO
The mass of CO needed to produce 208 kg of methanol is approximately 181.846 kg.
we need to use the stoichiometry of the balanced chemical reaction. The balanced reaction is[tex]\[ CO(g) + 2H_2(g) \rightarrow CH_3OH(g) \] \\[/tex]
From the stoichiometry of the reaction, we can see that 1 mole of CO reacts with 2 moles of H2 to produce 1 mole of CH3OH. The molar masses of CO and CH3OH are needed to convert from mass to moles and vice versa.
The molar mass of CH3OH is:
[tex]\[ M_{CH_3OH} = 12.01 \, \text{g/mol (C)} + 4 \times 1.008 \, \text{g/mol (H)} + 16.00 \, \text{g/mol (O)} = 32.042 \, \text{g/mol} \][/tex]
The molar mass of CO is:
[tex]\[ M_{CO} = 12.01 \, \text{g/mol (C)} + 16.00 \, \text{g/mol (O)} = 28.01 \, \text{g/mol} \][/tex]
Now, we convert the mass of methanol to moles of methanol:
[tex]\[ n_{CH_3OH} = \frac{m_{CH_3OH}}{M_{CH_3OH}} = \frac{208 \times 10^3 \, \text{g}}{32.042 \, \text{g/mol}} \][/tex]
Next, we use the stoichiometry of the reaction to find the moles of CO needed:
[tex]\[ n_{CO} = n_{CH_3OH} \][/tex]
Finally, we convert the moles of CO to the mass of CO:
[tex]\[ m_{CO} = n_{CO} \times M_{CO} \][/tex]
Let's calculate the mass of CO required:
[tex]\[ n_{CH_3OH} = \frac{208 \times 10^3 \, \text{g}}{32.042 \, \text{g/mol}} \approx 6493.5 \, \text{mol} \][/tex]
Since the stoichiometry is 1:1 for CO to CH3OH, we have:
[tex]\[ n_{CO} = 6493.5 \, \text{mol} \][/tex]
Now, we convert moles of CO to grams:
[tex]\[ m_{CO} = 6493.5 \, \text{mol} \times 28.01 \, \text{g/mol} \approx 181846 \, \text{g} \][/tex]
Converting grams to kilograms:
[tex]\[ m_{CO} = \frac{181846 \, \text{g}}{1000} \approx 181.846 \, \text{kg} \][/tex]
