How do stoichiometric ratios relate to molar volume of a gas?

Answer 1

They don’t.

When a gas is at a specific temperature and pressure, its molar volume remains constant.

Molar volumes, however, can be computed using molar ratios or stoichiometric ratios.

AN EXAMPLE

#"MgCO"_3"(s)" + "H"_2"SO"_4"(aq)" → "MgSO"_4"(aq)" + "H"_2"O(l)" + "CO"_2"(g)"#
A mass of 24.0 g of #"Mg"# produced 7.07 L of #"CO"_2# at 25.0 °C and a pressure of 0.987 bar. What is the molar volume of #"CO"_2# at STP (1 bar and 0 °C)?

Resolution

(a) We must calculate the moles of #"MgCO"_3# and then use the molar ratio to get the moles of #"CO"_2#
#24.0 color(red)(cancel(color(black)("g MgCO"_3))) × (1 color(red)(cancel(color(black)("mol MgCO"_3))))/(84.31 color(red)(cancel(color(black)("g MgCO"_3)))) × (1 "mol CO"_2)/(1 color(red)(cancel(color(black)("mol MgCO"_3)))) = "0.2847 mol CO"_2#
(b) Next, we use the combined gas laws to calculate the volume of the #"CO"_2# at STP.
#color(blue)(bar(ul(|color(white)(a/a)(P_1V_1)/(T_1) = (P_2V_2)/(T_2)color(white)(a/a)|)))" "#

This formula can be rearranged to provide

#V_2 = V_1 × (P_1)/(P_2) × (T_2)/(T_1)#
#P_1 = "0.987 bar"; V_1 = "7.15 L"; T_1 = "298.15 K"# #P_2 = "1 bar"; color(white)(mm)V_2 = ?;color(white)(mml) T_2 = "273.15 K"#
∴ #V_2 = "7.15 L" × (0.987 color(red)(cancel(color(black)("bar"))))/(1 color(red)(cancel(color(black)("bar")))) × (273.15 color(red)(cancel(color(black)("K"))))/(298.15 color(red)(cancel(color(black)("K")))) = "6.466 L"#

(c) The molar volume is then determined.

#"Molar volume" = "6.466 L"/"0.2847 mol" = "22.7 L/mol"#

At STP, the molar volume is 22.7 L.

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Answer 2

Stoichiometric ratios relate to the molar volume of a gas through the ideal gas law equation, which states that the volume of a gas is directly proportional to the number of moles of the gas at constant temperature and pressure. Therefore, when stoichiometric ratios are used to determine the amounts of reactants and products in a chemical reaction, they also determine the corresponding volumes of gases involved in the reaction, assuming ideal gas behavior.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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