Question

CO(g) + 2H₂(g) → CH₃OH(g)

A 56 kg sample of CO(g) and a 6.0 kg sample of H₂(g) are combined in a closed vessel, where they react according to the equation above.

After the reaction has proceeded to completion, how many moles of CH₃OH(g) have been produced?
a. 1.5 x 10³mol
b. 2.0 x 10³mol
c. 2.5 x 10³mol
d. 3.0 x 10³mol

Answer 1

After converting the given masses of CO and H₂ to moles, and identifying CO as the limiting reactant, we determine that 2000 moles of CH₃OH are produced, corresponding to option (b) 2.0 x 10³ mol.

Step-by-step explanation:

To find out how many moles of CH3OH were produced from the given amount of reactants, we need to perform stoichiometric calculations based on the balanced chemical equation CO(g) + 2H2(g) → CH3OH(g). First, we calculate the number of moles of each reactant using their respective molecular weights. The molar mass of CO is 28.01 g/mol, and for H2 it's 2.016 g/mol.

For CO: 56 kg CO x (1000 g/1 kg) x (1 mol CO/28.01 g CO) = 2000 moles CO.
For H2: 6.0 kg H2 x (1000 g/1 kg) x (1 mol H2/2.016 g H2) = 2976.19 moles H2.

Next, we determine the limiting reactant, which is CO since it requires twice the amount of H2 for the reaction and CO has fewer moles. Thus, all moles of CO will react completely, and according to the stoichiometry of the balanced equation, for every 1 mole of CO that reacts, 1 mole of CH3OH will form.

So, we have produced 2000 moles of CH3OH, which corresponds to answer option (b) 2.0 x 103 mol.