Question

Problem page liquid hexane ch3ch24ch3 will react with gaseous oxygen o2 to produce gaseous carbon dioxide co2 and gaseous water h2o . suppose 1.72 g of hexane is mixed with 1.8 g of oxygen. calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. be sure your answer has the correct number of significant digits.

Answer 1

Answer:

$m_{CO_2}=1.6gCO_2$

Explanation:

Hello,

In this case, hexane's combustion is stood for the following chemical reaction:

$C_6H_{14}+\frac{19}{2} O_2-->6CO_2+7H_2O$

Now, we need to identify the limiting reactant for which we compute both hexane's and oxygen's moles as shown below:

$n_{C_6H_{14}}=1.72g\frac{1molC_6H_{14}}{86gC_6H_{14}}=0.02molC_6H_{14}\\n_{O_2}=1.8g\frac{1molO_2}{32gO_2}=0.056molO_2$

Afterwards, we compute the moles of hexane that completely react with 0.056 moles of oxygen via the stoichiometry:

$n_{C_6H_{14}}^{reacting}=0.056molO_2\frac{1molC_6H_{14}}{\frac{19}{2}molO_2}=0.0059molC_6H_{14}$

Now, since 0.02 moles of hexane are available but just 0.0059 moles react, the hexane is in excess and the oxygen is the limiting reactant, thus, the maximum mass of carbon dioxide is computed as shown below with two significant figures (due to the significant figures of the oxygen's initial mass) by using the moles of oxygen as the limiting reactant:

$m_{CO_2}=0.056molO_2*\frac{6molCO_2}{\frac{19}{2}molO_2}*\frac{44gCO_2}{1molCO_2} \\m_{CO_2}=1.6gCO_2$

Best regards.

Answer 2

Hexane formula is C₆H₁₄
combustion reaction with O₂ is as follows;
C₆H₁₄ + 19/2O₂ ----> 6CO₂ + 7H₂O
to get whole number coefficients lets multiply by 2
2 C₆H₁₄ + 19 O₂ ----> 12 CO₂ + 14 H₂O
the stoichiometry of C₆H₁₄ to O₂ is 2 :19
2 moles of C₆H₁₄ reacts with 19 mol of O₂
both of these 2 reactants could be fully used up if the amounts are present in the correct stoichiometry or one reactant could be in excess and other could be the limiting reactant. Limiting reactant is when the reactant is fully consumed, products formed depends on this reactant.
amount of C₆H₁₄ moles - 1.72 g / 86.18 g/mol = 0.0200 mol
amount of O₂ moles - 1.8 g / 32 g/mol = 0.056 mol
if hexane is limiting reactant 
2:19 ratio , number of O₂ moles - 0.0200/2 *19 = 0.19
but 0.19 moles of O₂ aren't present, this means that O₂ is limiting reactant

stoichiometry of O₂ to CO₂ is 19:12
if 19 moles of O₂ is consumed - 12 mol of CO₂ produced 
then 1 mol of O₂ - gives 12/19 mol of CO₂
therefore 0.056 of O₂ - gives 12/19 * 0.056 mol 
                                  = 0.035 mol of CO₂
then mass of CO₂ = 0.035 mol * 12 g/mol 
                             = 0.42 g