Question

A synthesis reaction takes place when carbon monoxide (CO) and hydrogen gas (H2) react to form methanol (CH3OH). How many grams of methanol are produced when 2.8 grams of carbon monoxide reacts with 0.50 grams of hydrogen gas?
A. 3.2 grams
B. 3.6 grams
C. 4.0 grams
D. 6.0 grams

Answer 1

Answer:

A. 3.2grams

Explanation:

First we need to get the chemical equation for this reaction:

CO + H₂ → CH₃OH

We then need to balance the equation:

CO + 2H₂ → CH₃OH

Our next step is to first convert our given into moles. We do this by figuring out first how many grams of each substance there are in 1 mole by adding up the atomic mass of each element in each substance.

             Carbon(1)       Oxygen(1)

CO =       12.011(1)      +   15.999(1)   = 28.01 g/mole

            Hydrogen(2)

H₂ =        1.008(2)        =  2.016g/mole

We can then use this to determine how many moles of each reactant we have given the mass.

$CO = 2.8g\\2.8g\times\dfrac{1mole}{28.01g}=0.1moles\\\\H_{2}=0.50g\\\\0.50g\times\dfrac{1mole}{2.016g}=0.469moles$

So here we have our new given:

0.1 moles of CO

0.496 moles of H₂

We then need to determine how much product we produce with our given.

According to our chemical equation we can assume that:

For every 1 mole of CO we can produce 1 mole of CH₃OH

Fore every 2 moles of H₂ we can produce 1 mole of CH₃OH

Using this ratio we can determine how much product each reactant will produce by using the ratios:

$0.1 moles of CO\times\dfrac{1moleofCH_{3}OH}{1moleofCO}=0.1moles of CH_{3}OH\\\\ 0.496molesofH_{2}\times\dfrac{1moleofCH_{3}OH}{2molesofH_{2}}=0.248molesofCH_{3}OH$

Now notice that they are not equal. The reactant that produces the least amount of product will be our limiting reactant. The limiting reactant determines the amount of product that is produced, because once it is completely used up, the reaction stops. So in this case, the amount of CH₃OH that is produced is 0.10 moles.

Now we need to figure out how many this is in grams. To do that we need to find out how many grams of CH₃OH there are in 1 mole of CH₃OH.

                   Carbon (1)             Hydrogen(4)          Oxygen(1)

CH₃OH =      12.001(1)     +           1.008(4)       +       15.999(1)

            =      12.001         +           4.032          +        15.999     = 32.032g/mole

We then use this for converting:

$0.10 moles\times\dfrac{32.032g}{1mole}=3.2032g$

So the reaction will produce 3.2032g of CH₃OH or 3.2g of CH₃OH