2C31H64 + 63O2 -- 62CO2 + 64H2O Whatmass of water will be produced in the reaction? Mass before burning 0.72 g after burning 0.27 g

Question

2C31H64 + 63O2 --> 62CO2 + 64H2O Whatmass of water will be produced in the reaction? Mass before burning 0.72 g after burning 0.27 g

Wyatt Adkins · Accepted Answer

We write first the stoichiometric equation.........

#C_31H_64(s) + 47O_2(g) rarr 31CO_2(g) + 32H_2O(l)# Now #C_31H_64# is probably a wax or even a tar......Practically, we would expect incomplete combustion with such a hydrocarbon. If we assume complete combustion, then.............(and I assume, perhaps wrongly, that a #0.5*g# mass combusts) #"Moles of hydrocarbon"=(0.50*g)/(436.85*g*mol^-1)=1.14xx10^-3*mol#. And the equation CLEARLY indicates that #1.14xx10^-3*molxx32=3.66xx10^-2*mol# water result, i.e. a mass of #0.659*g#.

Kai Bowman · Answer

undefined To find the mass of water produced, you can first calculate the change in mass before and after the reaction:

Initial mass - Final mass = Change in mass

0.72 g - 0.27 g = 0.45 g

Since each mole of water has a mass of approximately 18 grams, you can use this information to find the moles of water produced:

0.45 g / 18 g/mol = 0.025 mol

Now, you can use the stoichiometry of the reaction to find the mass of water produced:

64 moles of water are produced for every 2 moles of $C_{31}H_{64}$.

So, for 0.025 moles of $C_{31}H_{64}$:

$$
\frac{{64 \times 0.025 \, \text{mol}}}{{2 \, \text{mol}}} = 0.8 \, \text{mol} \text{ of } H_2O
$$

Finally, to find the mass of water produced:

0.8 mol * 18 g/mol = 14.4 g

Therefore, 14.4 grams of water will be produced in the reaction.

2C31H64 + 63O2 --&gt; 62CO2 + 64H2O Whatmass of water will be produced in the reaction? Mass before burning 0.72 g after burning 0.27 g

2C31H64 + 63O2 --> 62CO2 + 64H2O Whatmass of water will be produced in the reaction? Mass before burning 0.72 g after burning 0.27 g