Question

Consider the equation for the combustion of methanol: CH3OH + 3O2 2CO2 + 4H2O. If 265g of CO2 is produced in this reaction, what mass of methanol was required? (assume excess oxygen)

Answer 1

this is chem but nobody was on that server

Answer 2

and i have a test tommorow

Answer 3

hello

Answer 4

The coefficients of each reactant/product represents the number of moles of the substance needed in the reaction. This means you'll have to convert from mass to moles, then back to mass.

Answer 5

\[\begin{array}{c|c}
\text{molecule}&\text{molar mass }\left(\frac{\text{g}}{\text{mol}}\right)\\
\hline
\text{CH}_3\text{OH}&12+4\times1+16=32\\
\text{O}_2&2\times16=32\\
\text{CO}_2&12+2\times16=88\\
\text{H}_2\text{O}&2\times1+16=18
\end{array}\]
So, if 256 g of \(\text{CO}_2\) were produced, this gives
\[256\text{ g CO}_2\times\frac{1\text{ mol CO}_2}{44\text{ g CO}_2}\times\frac{1\text{ mol CH}_3\text{OH}}{2\text{ mol CO}_2}\times\frac{32\text{ g CH}_3\text{OH}}{1\text{ mol CH}_3\text{OH}}=\cdots\text{ g CH}_3\text{OH}\]