You allow 75 grams of carbon monoxide to react with 58 gramsof hydrogen accordin

Question

You allow 75 grams of carbon monoxide to react with 58 gramsof hydrogen according to the following process. Determine thelimiting reactant and the amount of methanol produced. Co2 (g) + H2(g) equals CH3OH You allow 75 grams of carbon monoxide to react with 58 gramsof hydrogen according to the following process. Determine thelimiting reactant and the amount of methanol produced. Co2 (g) + H2(g) equals CH3OH

Explanation / Answer

this problem is a limiting reacent so by that you want to find the reactent that will create the least amount ofproduct. since the 1 to 1 to 1 ratio on the equation that means whicheverhas less moles of CO2 or H2 will be the limiting reactent so. wait the equation doesnt represent the words though ... CO + 2 H2 > CH3OH so lets use that equation. that means it takes twice as many moles to make methanol as justcarbon monoxide. so moles of CO 2.67moles H2 28.77 moles now compare which is bigger moles of CO or H2/2 obviously H2 is bigger so the lmiting reactent must be CO ths use your moles of CO then follow through the equation to showtaht 1 mol of CO will form 1 mol of product then multiply by molarmass of methanol so 85.8 grams of methanol will be formed

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