Chemistry, 11.06.2020 21:57 rlymyaa15

Problem PageQuestion Liquid hexane CH3CH24CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 60. g of hexane is mixed with 74.5 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to 3 significant digits.

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Answer from: PENTALIZARD2834

43.45g of water would be produced from the reaction.

Explanation:

Liquid became reacts with oxygen to produce carbon dioxide and water.

This type of reaction is known as combustion reaction between alkanes.

Equation of reaction.

Assuming the reaction occurs in an unlimited supply of oxygen,

2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O

From the above equation of reaction,

2 moles of C₆H₁₄ reacts with 19 moles of O₂ to produce 14 moles of H₂O.

To find the theoretical mass,

Number of moles = mass / molar mass

Molar mass of C₆H₁₄ = 86g/mol

Molar mass of O₂ = 16g/mol × 2 = 32g/mol

Molar mass of H₂O = 18g/mol

Mass of H₂O = number of moles × molar mass

Mass of H₂O = 14 × 18 = 252g

Mass of C₆H₁₄ = number of moles × molar mass

Mass of C₆H₁₄ = 2 × 86 = 172g

Mass of O₂ = number of moles × molar mass

Mass of O₂ = 19 × 32 = 608g

From the equation of reaction,

172g of C₆H₁₄ reacts with 608g of O₂ to produce 252g of H₂O

(172 + 608)g of reactants produce 252g of H₂O

780g of reactants produce 252g of H₂O

(60 + 75.5)g of reactants will produce a x g of H₂O

780g of reactants = 252g of H₂O

134.5g of reactants = x g of H₂O

X = (134.5 × 252) / 780

X = 43.45g of H₂O

Therefore, 43.45g of H₂O would be produced from 60g of hexane and 74.5g of oxygen

Answer from: trinati6965

$m_{H_2O}=30.9gH_2O$

Explanation:

Hello,

In this case, the combustion of hexane is given by:

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

The next step is to compute the reacting moles of hexane:

$n_{C_6H_{14}}=60gC_6H_{14}*\frac{1molC_6H_{14}}{86gC_6H_{14}} =0.698molC_6H_{14}$

Then, the moles of hexane consumed by 74.5 g of oxygen using the molar ratio in the chemical reaction (1:19/2):

$n_{C_6H_{14}}=74.5gO_2*\frac{1molO_2}{32gO_2} *\frac{1molC_6H_{14}}{19/2molO_2} =0.245molC_6H_{14}$

Therefore, as less moles of hexane are consumed by oxygen, it is in excess, so we compute the mass of water produced by the consumed 0.245 moles of hexane:

$m_{H_2O}=0.245molC_6H_{14}*\frac{7molH_2O}{1molC_6H_{14}}*\frac{18gH_2O}{1molH_2O} \\\\m_{H_2O}=30.9gH_2O$

Best regards.

Answer from: Quest

the dependant variable is spilled water

the independent is the heat

Answer from: Quest

elements make up everything, and they combine to make compounds.

the answer is c

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Problem PageQuestion Liquid hexane CH3CH24CH3 will react with gaseous oxygen O2 to produce gaseous c...

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