Consider the unbalanced equation: $C_{6} H_{14} + O_{2} \to C O_{2} + H_{2} O$ $C_6H_14 + O_2 -> CO_2 + H_2O$ . What mass of $O_{2}$ $O_2$ is required to react with $11.5 g$ $"11.5 g"$ of $C_{6} H_{14}$ $C_6H_14$ ?

Question

Consider the unbalanced equation: $C_{6} H_{14} + O_{2} \to C O_{2} + H_{2} O$ $C_6H_14 + O_2 -> CO_2 + H_2O$ . What mass of $O_{2}$ $O_2$ is required to react with $11.5 g$ $"11.5 g"$ of $C_{6} H_{14}$ $C_6H_14$ ?

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Answer 1 · 2017-03-15T05:27:22

${0.0400 g O}_{2}$ $"0.0400 g O"_2"$ is required to react with $11.5 g$ $"11.5 g"$ of $C_{6} H_{14}$ $"C"_6"H"_14"$ .

Explanation:

Start with a balanced equation.

${2C}_{6} H_{14} + {19O}_{2}$ $"2C"_6"H"_14 + "19O"_2$ $\to$ $rarr$ ${12CO}_{2} + {14H}_{2} O$ $"12CO"_2 + "14H"_2"O"$

Use the balanced equation to determine the mole ratios between $C_{6} H_{14}$ $"C"_6"H"_14"$ and $O_{2}$ $"O"_2"$ .

$\frac{{2 mol C}_{6} H_{14}}{{19 mol O}_{2}}$ $("2 mol C"_6"H"_14)/("19 mol O"_2")$ and $\frac{{19 mol O}_{2}}{{2 mol C}_{6} H_{14}}$ $("19 mol O"_2)/("2 mol C"_6"H"_14)$

Determine the molar masses of $C_{6} H_{14}$ $"C"_6"H"_14$ and $O_{2}$ $"O"_2"$ .

$C_{6} H_{14} :$ $"C"_6"H"_14:$ $86.178 g/mol$ $"86.178 g/mol"$
https://www.ncbi.nlm.nih.gov/pccompound?term=C6H14
$O_{2} :$ $"O"_2:$ $31.998 g/mol$ $"31.998 g/mol"$

Determine the moles of $C_{6} H_{14}$ $"C"_6"H"_14"$ by dividing its given mass by its molar mass.

$(11.5 cancel"g C"_6"H"_14)/(86.178cancel"g"/"mol")="0.13344 mol C"_6"H"_14"$

Determine moles of $"O"_2"$ by multiplying mol $"C"_6"H"_14"$ by the mol ratio that has $"O"_2"$ in the numerator.

$0.13344cancel"mol C"_6"H"_14xx(19"mol O"_2)/(2 cancel"mol C"_6"H"_14)="1.2678 mol O"_2"$

Determine the mass of $"O"_2"$ that will react with $11.5"g C"_6"H"_14"$ by multiplying the moles $"O"_2"$ by its molar mass.

$1.2678cancel"mol O"_2xx(31.998"g O"_2)/(1cancel"mol O"_2)="0.0400 g O"_2"$ (rounded to three significant figures)

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Consider the unbalanced equation: $C_{6} H_{14} + O_{2} \to C O_{2} + H_{2} O$ $C_6H_14 + O_2 -> CO_2 + H_2O$ . What mass of $O_{2}$ $O_2$ is required to react with $11.5 g$ $"11.5 g"$ of $C_{6} H_{14}$ $C_6H_14$ ?

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Explanation:

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Consider the unbalanced equation: C_6H_14 + O_2 -> CO_2 + H_2OC6H14+O2→CO2+H2OC_6H_14 + O_2 -> CO_2 + H_2O. What mass of O_2O2O_2 is required to react with "11.5 g"11.5 g"11.5 g" of C_6H_14C6H14C_6H_14?

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Consider the unbalanced equation: $C_{6} H_{14} + O_{2} \to C O_{2} + H_{2} O$ $C_6H_14 + O_2 -> CO_2 + H_2O$ . What mass of $O_{2}$ $O_2$ is required to react with $11.5 g$ $"11.5 g"$ of $C_{6} H_{14}$ $C_6H_14$ ?