According to the problem statement, if you did everything right,
$$\begin{align}
\text{AFR} &= 37.54 \\
\text{AFR}_{\text{stoich}} &= \frac{\nu_{\text{oxygen}}}{\nu_{\text{fuel}}}\times 4.76 \\
\ce{C4H10 + $6.5$\,O2 &-> 4 CO2 + 5 H2O} \\
\text{AFR}_{\text{stoich}} &= 4.76 \times 6.5 = 30.94 \\
\text{Equivalence ratio} &= \frac {37.54}{30.94}=1.21
\end{align}$$
Now there are two types of equivalence ratio. One is air to fuel and another is fuel to air. I suspect in the answer they meant fuel to air equivalence ratio:
$$\text{FAR equivalence ratio} = \frac{1}{\text{AFR equivalence ratio}} = \frac{1}{1.21}=0.82$$
According to the problem statement, if you did everything right,$$\begin{align}\text{AFR} &= 37.54 \\\text{AFR}_{\text{stoich}} &= \frac{\nu_{\text{oxygen}}}{\nu_{\text{fuel}}}\times 4.76 \\\ce{C4H10 + $6.5$\,O2 &-> 4 CO2 + 5 H2O} \\\text{AFR}_{\text{stoich}} &= 4.76 \times 6.5 = 30.94 \\\text{Equivalence ratio} &= \frac {37.54}{30.94}=1.21\end{align}$$Now there are two types of equivalence ratio. One is air to fuel and another is fuel to air. I suspect in the answer they meant fuel to air equivalence ratio: $$\text{FAR equivalence ratio} = \frac{1}{\text{AFR equivalence ratio}} = \frac{1}{1.21}=0.82$$
I followed this procedure. Note that "oxygen" here refers to ${\rm O}_{2}$.
They define the equivalence ratio as: $${\rm F/A\over (F/A)_s}$$ where F/A is your fuel to oxygen ratio (moles of fuel divided by moles of oxygen), and (F/A)$_{\rm s}$ is the "stoichiometric ratio" defined as the number of moles of oxygen needed to combust 1 mole of fuel.
For the numerator, you have 4/58 moles of fuel (F) and [0.21 $\times$ (75/29)] moles of oxygen (A) (0.21 because 21% of air is oxygen). Your numerator (F/A) is equal to 0.126.
For the denominator (F/A)$_{\rm s}$, we know that for each mole of fuel ${\rm C_mH_n}$ you need ${\rm (m + n/4)}$ moles of oxygen (see linked text for that). For butane (your fuel) that means you need 6.5 moles of oxygen to combust 1 mole of fuel. Your denominator is 0.153 (that's just 1/6.5).
I followed this procedure. Note that "oxygen" here refers to ${\rm O}_{2}$.
They define the equivalence ratio as: $${\rm F/A\over (F/A)_s}$$ where F/A is your fuel to oxygen ratio (moles of fuel divided by moles of oxygen), and (F/A)$_{\rm s}$ is the "stoichiometric ratio" defined as the number of moles of oxygen needed to combust 1 mole of fuel.
For the numerator, you have 4/58 moles of fuel (F) and [0.21 $\times$ (75/29)] moles of oxygen (A) (0.21 because 21% of air is oxygen). Your numerator (F/A) is equal to 0.126.
For the denominator (F/A)$_{\rm s}$, we know that for each mole of fuel ${\rm C_mH_n}$ you need ${\rm (m + n/4)}$ moles of oxygen (see linked text for that). For butane (your fuel) that means you need 6.5 moles of oxygen to combust 1 mole of fuel. Your denominator is 0.153 (that's just 1/6.5).
$$\begin{align} \text{AFR (Air to Fuel ratio)} &= \frac{n_{\text{air}}}{n_{\text{fuel}}} \\ \text{Equivalence ratio} &= \frac{\text{AFR}}{\text{AFR}_{\text{stoich}}} \\ \end{align}$$
According to the problem statement, if you did everything right, $$\begin{align} \text{AFR} &= 37.54 \\ \text{AFR}_{\text{stoich}} &= \frac{\nu_{\text{oxygen}}}{\nu_{\text{fuel}}}\times 4.76 \\ \ce{C4H10 + $6.5$\,O2 &-> 4 CO2 + 5 H2O} \\ \text{AFR}_{\text{stoich}} &= 4.76 \times 6.5 = 30.94 \\ \text{Equivalence ratio} &= \frac {37.54}{30.94}=1.21 \end{align}$$ Now there are two types of equivalence ratio. One is air to fuel and another is fuel to air. I suspect in the answer they meant fuel to air equivalence ratio: $$\text{FAR equivalence ratio} = \frac{1}{\text{AFR equivalence ratio}} = \frac{1}{1.21}=0.82$$
$$\begin{align}\text{AFR (Air to Fuel ratio)} &= \frac{n_{\text{air}}}{n_{\text{fuel}}} \\\text{Equivalence ratio} &= \frac{\text{AFR}}{\text{AFR}_{\text{stoich}}} \\\end{align}$$
According to the problem statement, if you did everything right,$$\begin{align}\text{AFR} &= 37.54 \\\text{AFR}_{\text{stoich}} &= \frac{\nu_{\text{oxygen}}}{\nu_{\text{fuel}}}\times 4.76 \\\ce{C4H10 + $6.5$\,O2 &-> 4 CO2 + 5 H2O} \\\text{AFR}_{\text{stoich}} &= 4.76 \times 6.5 = 30.94 \\\text{Equivalence ratio} &= \frac {37.54}{30.94}=1.21\end{align}$$Now there are two types of equivalence ratio. One is air to fuel and another is fuel to air. I suspect in the answer they meant fuel to air equivalence ratio: $$\text{FAR equivalence ratio} = \frac{1}{\text{AFR equivalence ratio}} = \frac{1}{1.21}=0.82$$
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I followed this procedure. Note that "oxygen" here refers to ${\rm O}_{2}$.
They define the equivalence ratio as: $${\rm F/A\over (F/A)_s}$$ where F/A is your fuel to oxygen ratio (moles of fuel divided by moles of oxygen), and (F/A)$_{\rm s}$ is the "stoichiometric ratio" defined as the number of moles of oxygen needed to combust 1 mole of fuel.
For the numerator, you have 4/58 moles of fuel (F) and [0.21 $\times$ (75/29)] moles of oxygen (A) (0.21 because 21% of air is oxygen). Your numerator (F/A) is equal to 0.126.
For the denominator (F/A)$_{\rm s}$, we know that for each mole of fuel ${\rm C_mH_n}$ you need ${\rm (m + n/4)}$ moles of oxygen (see linked text for that). For butane (your fuel) that means you need 6.5 moles of oxygen to combust 1 mole of fuel. Your denominator is 0.153 (that's just 1/6.5).
Dividing 0.126/0.153 = 0.82
I followed this procedure. Note that "oxygen" here refers to ${\rm O}_{2}$.
They define the equivalence ratio as: $${\rm F/A\over (F/A)_s}$$ where F/A is your fuel to oxygen ratio (moles of fuel divided by moles of oxygen), and (F/A)$_{\rm s}$ is the "stoichiometric ratio" defined as the number of moles of oxygen needed to combust 1 mole of fuel.
For the numerator, you have 4/58 moles of fuel (F) and [0.21 $\times$ (75/29)] moles of oxygen (A) (0.21 because 21% of air is oxygen). Your numerator (F/A) is equal to 0.126.
For the denominator (F/A)$_{\rm s}$, we know that for each mole of fuel ${\rm C_mH_n}$ you need ${\rm (m + n/4)}$ moles of oxygen (see linked text for that). For butane (your fuel) that means you need 6.5 moles of oxygen to combust 1 mole of fuel. Your denominator is 0.153 (that's just 1/6.5).
Dividing 0.126/0.153 = 0.82
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