Question #1e2a5

1 Answer
Jun 17, 2016

Answer:

#"15.7 kg CH"_3"OH"#

Explanation:

Your starting point here will be the balanced chemical equation that describes this synthesis reaction

#"CO"_ ((g)) + color(red)(2)"H"_ (2(g)) -> "CH"_ 3"OH"_ ((g))#

Notice that the reaction consumes carbon monoxide, #"CO"#, and hydrogen gas, #"H"_2#, in a #1:color(red)(2)# mole ratio. Keep this in mind.

Now, the problem provides you with kilograms of carbon monoxide and of hydrogen gas. Convert these to grams first

#13.7 color(red)(cancel(color(black)("kg"))) * (10^3"g")/(1color(red)(cancel(color(black)("kg")))) = 1.37 * 10^4"g"#

#17.2color(red)(cancel(color(black)("kg"))) * (10^3"g")/(1color(red)(cancel(color(black)("kg")))) = 1.72 * 10^4"g"#

Use the molar masses of the two reactants to convert these to moles

#1.37 * 10^4 color(red)(cancel(color(black)("g"))) * "1 mole CO"/(28.01color(red)(cancel(color(black)("g")))) = 4.891 * 10^2"moles CO"#

#1.72 * 10^4color(red)(cancel(color(black)("g"))) * "1 mole H"_2/(2.016color(red)(cancel(color(black)("g")))) = 8.532 * 10^3"moles H"_2#

Your next goal is to determine whether or not you're dealing with a limiting reagent. Pick a reactant and use the aforementioned mole ratio to see if you have enough of the second reactant to allow for all the moles of the first reactant to take part in the reaction

#4.891 * 10^2 color(red)(cancel(color(black)("moles CO"))) * (color(red)(2)color(white)(a)"moles H"_2)/(1color(red)(cancel(color(black)("mole CO")))) = 9.782 * 10^2"moles H"_2#

As you can see, you have more hydrogen gas than you need. This means that carbon monoxide will act as alimiting reagent, i.e. it will be consumed before all the moles of hydrogen gas will get the chance to react.

In other words, hydrogen gas is in excess.

Notice that you have a #1:1# mole ratio between carbon monoxide and methanol. This tells you that the reaction will produce as many moles of methanol as you have moles of carbon monoxide that take part in the reaction.

You will thus have

#4.891 * 10^2 color(red)(cancel(color(black)("moles CO"))) * ("1 mole CH"_3"OH")/(1color(red)(cancel(color(black)("mole CO")))) = 4.891 * 10^2"moles CH"_3"OH"#

This represents the reaction's theoretical yield, which essentially tells you how much product you can expect for a #100%# conversion rate.

If all the moles of carbon monoxide and all the moles of hydrogen gas that take part in the reaction produce methanol, you will end up with #4.891 * 10^2# moles of product.

To express this in grams, use the compound's molar mass

#4.891 * 10^2 color(red)(cancel(color(black)("moles CH"_3"OH"))) * "32.042 g"/(1color(red)(cancel(color(black)("mole CH"_3"OH")))) = 1.567 * 10^4"g"#

Finally, expressed in kilograms and rounded to three sig figs, the answer will be

#"theoretical yield" = color(green)(|bar(ul(color(white)(a/a)color(black)("15.7 kg")color(white)(a/a)|)))#