Consider this reaction:
[tex]2 Mg(s)+O_2(g) \rightarrow 2 MgO(s)[/tex]
What volume (in milliliters) of oxygen gas is required to react with 4.03 g of Mg at STP?
A. 1860 mL
B. 2880 mL
C. 3710 mL
D. [tex]45,100 dmL[/tex]
Answer (2)
To react with 4.03 g of magnesium, approximately 1858 mL of oxygen gas is required at STP. The closest answer from the options is 1860 mL. Thus, the answer is A. 1860 mL.
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Calculate the moles of Mg: m o l e s M g = 24.305 g / m o l 4.03 g = 0.1658 m o l .
Determine the moles of O 2 : m o l e s O 2 = 2 1 × 0.1658 m o l = 0.0829 m o l .
Use the ideal gas law to find the volume of O 2 in liters: V = 1 a t m 0.0829 m o l × 0.0821 m o l ⋅ K L ⋅ a t m × 273.15 K = 1.859 L .
Convert the volume to milliliters: V o l u m e O 2 = 1.859 L × 1000 = 1859 m L . The final answer is 1859 m L .
Explanation
Problem Analysis We are given the reaction 2 M g ( s ) + O 2 ( g ) → 2 M g O ( s ) and asked to find the volume of oxygen gas required to react with 4.03 g of Mg at STP.
Calculate Moles of Mg First, we need to calculate the number of moles of Mg. The molar mass of Mg is 24.305 g/mol. So, the number of moles of Mg is given by: m o l e s M g = m o l a r ma s s M g ma s s M g = 24.305 g / m o l 4.03 g = 0.1658 m o l
Calculate Moles of O2 From the balanced chemical equation, we see that 2 moles of Mg react with 1 mole of O 2 . Therefore, the number of moles of O 2 required is half the number of moles of Mg: m o l e s O 2 = 2 1 × m o l e s M g = 2 1 × 0.1658 m o l = 0.0829 m o l
Calculate Volume of O2 in Liters Now, we use the ideal gas law to find the volume of O 2 at STP. At STP, the temperature T = 273.15 K and the pressure P = 1 atm. The ideal gas constant R = 0.0821 L atm / (mol K). The volume of O 2 in liters is given by: V = P n RT = 1 a t m 0.0829 m o l × 0.0821 m o l ⋅ K L ⋅ a t m × 273.15 K = 1.859 L
Calculate Volume of O2 in Milliliters Finally, we convert the volume of O 2 from liters to milliliters: V o l u m e O 2 ( m L ) = V o l u m e O 2 ( L ) × 1000 = 1.859 L × 1000 = 1859 m L Therefore, the volume of oxygen gas required to react with 4.03 g of Mg at STP is approximately 1859 mL.
Final Answer The volume of oxygen gas required to react with 4.03 g of Mg at STP is approximately 1859 mL.
Examples
In the field of metallurgy, calculating the precise amount of oxygen needed to react with a metal like magnesium is crucial for producing metal oxides with specific properties. For example, magnesium oxide (MgO) is used in various applications, including refractory materials, electrical insulators, and pharmaceuticals. Knowing the stoichiometry of the reaction and the ideal gas law allows engineers to control the reaction conditions and produce MgO with the desired purity and characteristics. This ensures the final product meets the required specifications for its intended use.
