Question #0ea44

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Question #0ea44

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Answer 1 · 2015-11-01T11:58:50

$1.92 g$ $"1.92 g"$

Explanation:

Notice that the problem gives you three important parameters to work with

the volume occupied by the gas

the pressure of the gas

the temperature of the gas

This means that you can use the ideal gas law equation

$P V = n R T$ $color(blue)(PV = nRT)$

to find the number of moles of gas found in the sample. Once you know how many moes of carbon dioxide you have, you can use its molar mass to determine how many grams you have.

So, the ideal gas law equation establishes a relationship between the pressure of the gas, $P$ $P$ , and its volume, $V$ $V$ , on one hand, and the number of moles of gas, $n$ $n$ , and the temperature of the gas, $T$ $T$ , on the other.

Here $R$ $R$ is the universal gas constant and is usually given as

$R = 0.082 \frac{atm \cdot L}{mol \cdot K}$ $R = 0.082("atm" * "L")/("mol" * "K")$

This means that you must make sure that you're using the correct units for pressure, temperature, and volume, i.e. the units that match those used for $R$ $R$ .

In your case, that means converting the volume of the gas from mililiters to liters and the temperatue of the gas from degrees Celsius to Kelvin

$450.6 mL \cdot \frac{1 L}{10^{3} mL} = 0.4506 L$ $450.6color(red)(cancel(color(black)("mL"))) * "1 L"/(10^3color(red)(cancel(color(black)("mL")))) = "0.4506 L"$

and

$- {50.5}^{\circ} C + 273.15 = 226.65 K$ $-50.5^@"C" + 273.15 = "226.65 K"$

Now plug in your values and solve the ideal gas law equation for $n$ $n$

$n = \frac{P V}{R T}$ $n = (PV)/(RT)$

$n = \frac{1.80 atm \cdot 0.4506 L}{0.082 \frac{atm \cdot L}{mol \cdot K} \cdot 226.65 K} = 0.04364 moles$ $n = (1.80color(red)(cancel(color(black)("atm"))) * 0.4506color(red)(cancel(color(black)("L"))))/(0.082(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * 226.65color(red)(cancel(color(black)("K")))) = "0.04364 moles"$

Finally, you know that carbon dioxide's molar mass is equal to $44.01 g/mol$ $"44.01 g/mol"$ . This means that every mole of ${CO}_{2}$ $"CO"_2$ has a mass of $44.01 g$ $"44.01 g"$ .

In your case, the mass of the sample will be

$0.04364 moles \cdot \frac{44.01 g}{1 mole} = 1.92 g$ $0.04364color(red)(cancel(color(black)("moles"))) * "44.01 g"/(1color(red)(cancel(color(black)("mole")))) = color(green)("1.92 g")$

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Question #0ea44

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