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Chapter 5: Problem 52

A person accidentally swallows a drop of liquid oxygen, \(\mathrm{O}_{2}(l)\), which has a density of \(1.149 \mathrm{~g} / \mathrm{mL}\). Assuming the drop has a volume of \(0.050 \mathrm{~mL}\), what volume of gas will be produced in the person's stomach at body temperature \(\left(37^{\circ} \mathrm{C}\right)\) and a pressure of \(1.0 \mathrm{~atm}\) ?

Short Answer

Expert verified
The volume of O鈧� gas produced in the person's stomach at body temperature (37掳C) and a pressure of 1.0 atm will be 45.7 mL.

Step by step solution

01

Finding the mass of liquid oxygen drop

We are given the density of the liquid oxygen drop, which is 1.149 g/mL, and its volume, which is 0.050 mL. We will use these values to find the mass of this drop. Mass = Density 脳 Volume.
02

Calculate the mass of the liquid oxygen

We plug the values we have into the formula from Step 1: \( Mass = (1.149 \frac{g}{mL}) \times (0.050 mL) \) By multiplying the values, we get: \( Mass = 0.05745 \ g \)
03

Finding the number of moles of O鈧�

Now that we have the mass of the liquid oxygen, we will find the number of moles of O鈧�. We know the molar mass of O鈧� is 32.00 g/mol. We will use the formula: Moles of O鈧� = (mass of O鈧�) / (molar mass of O鈧�)
04

Calculate the number of moles of O鈧�

Plugging in the values from Step 2 and the molar mass of O鈧�, we get: \( Moles \ of \ O_{2} = \frac{0.05745 \ g}{32.00 \frac{g}{mol}} \) By dividing the mass by molar mass, we get: \( Moles \ of \ O_{2} = 0.0018 \ mol \)
05

Find the volume of O鈧� gas using the ideal gas law

We have the number of moles of O鈧� gas and we are given the pressure (1.0 atm) and temperature (37掳C or 310 K). The ideal gas law formula is: \(PV = nRT\) Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L atm /(K mol)), and T is the temperature in Kelvin. We will solve for V (volume of gas).
06

Calculate the volume of O鈧� gas

Plug in the values we have for P, n, R, and T, and solve for V: \( V = \frac{nRT}{P} \\ V = \frac{(0.0018 \ mol)(0.0821 \frac{L \cdot atm}{K \cdot mol})(310 \ K)}{1.0 \ atm} \) By multiplying and dividing the values, we get: \( V = 0.0457 \ L \)
07

Express the result in mL

We will convert the volume of O鈧� gas from liters to milliliters by multiplying by 1000: Volume of O鈧� gas = 0.0457 L 脳 1000 mL/L Volume of O鈧� gas = 45.7 mL Thus, the volume of O鈧� gas produced in the person's stomach at body temperature (37掳C) and a pressure of 1.0 atm will be 45.7 mL.

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Most popular questions from this chapter

Silane, \(\mathrm{SiH}_{4}\), is the silicon analogue of methane, \(\mathrm{CH}_{4}\). It is prepared industrially according to the following equations: $$ \begin{aligned} \mathrm{Si}(s)+3 \mathrm{HCl}(g) & \longrightarrow \mathrm{HSiCl}_{3}(l)+\mathrm{H}_{2}(g) \\ 4 \mathrm{HSiCl}_{3}(l) & \longrightarrow \mathrm{SiH}_{4}(g)+3 \mathrm{SiCl}_{4}(l) \end{aligned} $$ a. If \(156 \mathrm{~mL} \mathrm{HSiCl}_{3}(d=1.34 \mathrm{~g} / \mathrm{mL})\) is isolated when \(15.0 \mathrm{~L}\) \(\mathrm{HCl}\) at \(10.0 \mathrm{~atm}\) and \(35^{\circ} \mathrm{C}\) is used, what is the percent yield of \(\mathrm{HSiCl}_{3} ?\) b. When \(156 \mathrm{~mL} \mathrm{HSiCl}_{3}\) is heated, what volume of \(\mathrm{SiH}_{4}\) at \(10.0\) atm and \(35^{\circ} \mathrm{C}\) will be obtained if the percent yield of the reaction is \(93.1 \%\) ?

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