# Question #315e6

May 28, 2017

$14.33 \text{g of } {H}_{2} O$

#### Explanation:

${C}_{6} {H}_{12} {O}_{6} + 6 {O}_{2} \rightarrow 6 {H}_{2} O + 6 C {0}_{2}$

We start Dimensional analysis (the factor-label method) by writing the given information over 1:

$\frac{35 \text{g of } C {O}_{2}}{1}$

We look up the molar mass of $C {O}_{2}$ and multiply by that factor so that it converts the units to moles:

$\left(35 \text{g of "CO_2)/1(1"mol of "CO_2)/(44.01"g of } C {O}_{2}\right)$

Please observe how the units cancel and we are left with moles of $C {O}_{2}$:

$\left(35 \textcolor{red}{\cancel{\text{g of "CO_2)))/1(1"mol of "CO_2)/(44.01color(red)(cancel("g of } C {O}_{2}}}\right)$

We obtain the factor that 6 moles of water are produced when 6 moles of $C {O}_{2}$ are produced, from the equation:

$\left(35 \textcolor{red}{\cancel{\text{g of "CO_2)))/1(1color(red)cancel(("mol of "CO_2)))/(44.01color(red)(cancel("g of "CO_2)))(6"mol of "H_2O)/(6color(red)(cancel("mol of } C {O}_{2}}}\right)$

Look up the molar mass of water:

$\left(35 \textcolor{red}{\cancel{\text{g of "CO_2)))/1(1color(red)cancel(("mol of "CO_2)))/(44.01color(red)(cancel("g of "CO_2)))(6color(red)cancel(("mol of "H_2O)))/(6color(red)(cancel("mol of "CO_2)))(18.02"g of "H_2O)/(1color(red)cancel(("mol of } {H}_{2} O}}\right)$

We have the units in grams of ${H}_{2} O$; now, just do the multiplication or division:

$\frac{35}{44.01} \left(18.02\right) = 14.33 \text{g of } {H}_{2} O$