mass A moles A moles B particles B Referring to the equation above, how many grams of CH3OH must react in order to produce 17.5 g of H2O

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10-11-2020
Chemistry

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mass A moles A moles B particles B Referring to the equation above, how many grams of CH3OH must react in order to produce 17.5 g of H2O

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etiennevalentin etiennevalentin · Answer 1 · 2020-11-10T09:24:49+01:00

Answer:

17.5 g H2O x 1 mol H2O / 18.01 g H2O x 2 mol CH3OH / 4 mol H2O x 32.04 g CH3OH / 1 mol CH3OH = 15.5 g CH3OH

ANSWER: 15.5 g CH3OH

Explanation:

Hi Bash!

I believe we're in the same class because I was trying to understand this as well.

You have a missing piece of information which is the chemical equation the question is derived from. At the top of the study guide, you will find that the equation is: 2 CH3OH + 3 O2 --> 2 CO2 + 4 H2O

The given is 17.5 g of H2O and you're trying to find how many grams of CH3OH must react.

You'll begin by converting the grams of H2O into moles: 17.5 g H2O x 1 mol H2O/18.01 g H2O (18.01 is the molar mass and each mole has a mass of 18.01 g)

Then you will convert the moles of H2O into moles of CH3OH. There are 2 moles of CH3OH as seen in the reactant and there are 4 moles of H2O as seen in the product side : 2 mol CH3OH / 4 mol H2O x 32.04 g CH3OH / 1 mol CH3OH.

Combine all these steps together and you will get: 17.5 g H2O x 1 mol H2O / 18.01 g H2O x 2 mol CH3OH / 4 mol H2O x 32.04 g CH3OH / 1 mol CH3OH = 15.5 g CH3OH