The mass of methanol produced is 8.0 g.
We have the masses of two reactants, so this is a <em>limiting reactant</em> problem.
We know that we will need a <em>balanced equation</em> with masses, moles, and molar masses of the compounds involved.
<em>Step 1</em>. <em>Gather all the informatio</em>n in one place with molar masses above the formulas and everything else below them.
MM: ___28.01 2.016 ___32.04
_______CO + 2H_2 → CH_3OH
Mass/g: 7.0 __2.5
<em>Step 2</em>. Calculate the <em>moles of each reactant</em>
Moles of CO = 7.0 g CO × (1 mol CO/28.01g CO) = 0.250 mol CO
Moles of H_2 =2.5 g H_2 × (1 mol H_2/2.016 g H_2) = 1.24 mol H_2
<em>Step 3. </em>Identify the<em> limiting reactan</em>t
Calculate the <em>moles of CH_3OH</em> we can obtain from each reactant.
<em>From CO</em>: Moles of CH_3OH = 0.250 mol CO × (1 mol CH_3OH /1 mol CO)
= 0.250 mol CH_3OH
<em>From H_2</em>: Moles of CH_3OH = 1.24 mol H_2 × (1 mol CH_3OH /2 mol H_2)
= 0.620 mol CH_3OH
<em>CO is the limiting reactant</em> because it gives the smaller amount of CH_3OH.
<em>Step 4</em>. Calculate the <em>mass of CH_3OH</em>
Mass of CH_3OH = 0.250 mol CH_3OH × (32.04 g CH_3OH /1 mol CH_3OH) = 8.0 g CH_3OH
