Chemistry Stoichiometric Problems~~

Question

Liquid methanol, CH3OH, can be produced through the reaction of carbon monoxide and hydrogen gas in the presence of a catalyst.
a) If 350.0 mL of carbon monoxide is mixed with 650.0 mL of hydrogen in sealed reaction vessel @ STP, what mass of liquid methanol will be formed?
b) Given that the density of liquid methanol is 0.7914 g/mL, what volume would the product occupy?

Answer 1

a)
If 350.0 mL of carbon monoxide is mixed with 650.0 mL of hydrogen in sealed reaction vessel @ STP, what mass of liquid methanol will be formed?

CO(s) + 2H2(g) → CH3OH(l)
Mole ratio CO : H2 = 1 : 2
No. of moles of CO added = volume / (molar volume) = 350.0/22400 = 0.01563 mol
No. of moles of H2 added = volume / (molar volume) = 650.0/22400 = 0.02902 mol
To complete react 0.01563 mol of CO, 0.03124 mol (> 0.02902 mol) of H2 is needed. Hence, H2 is the limiting reactant.

Mole ratio H2 : CH­­3OH = 2 : 1
No. of moles of CH3OH formed = 0.02902 x (1/2) = 0.01451 mol
Molar mass of CH3OH = 12 + 1x4 + 16 = 32 g mol-1
Mass of CH3OH = mol x (molar mass) = 0.01451 x 32 = 0.4643 g

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b)
Given that the density of liquid methanol is 0.7914 g/mL, what volume would the product occupy?

Volume of CH3OH formed = (mass / density) = (0.4643 / 0.7914) = 0.5867 mL