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. 1960 mL
B. 2880 mL
C. 3710 mL
D. [tex]45,100 mL[/tex]
Answer (2)
The volume of oxygen gas required to react with 4.03 g of magnesium at STP is approximately 1859 mL. The closest option is 1960 mL. Thus, the correct answer is A. 1960 mL.
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Calculate the number of moles of Mg: n M g = 24.305 4.03 = 0.1658 mol.
Determine the number of moles of O 2 : n O 2 = 2 1 n M g = 0.0829 mol.
Use the ideal gas law to find the volume of O 2 : V O 2 = P n O 2 RT = 1 ( 0.0829 ) ( 0.0821 ) ( 273.15 ) = 1.859 L.
Convert the volume to milliliters: V O 2 = 1.859 L ∗ 1000 = 1859 mL. Therefore, the closest answer is 1960 m L .
Explanation
Calculate moles of Mg First, we need to find the number of moles of Mg. We are given that the mass of Mg is 4.03 g. The molar mass of Mg is approximately 24.305 g/mol. We can calculate the number of moles of Mg using the formula: n M g = M o l a r M a s s M g ma s s M g n M g = 24.305 g / m o l 4.03 g = 0.1658 m o l
Calculate moles of O2 Next, we need to determine the number of moles of O 2 required to react with the Mg. From the balanced chemical equation, we have: 2 M g ( s ) + O 2 ( g ) → 2 M g O ( s ) This tells us that 2 moles of Mg react with 1 mole of O 2 . Therefore, the number of moles of O 2 is half the number of moles of Mg: n O 2 = 2 1 n M g n O 2 = 2 1 ( 0.1658 m o l ) = 0.0829 m o l
Calculate volume of O2 in liters Now, we can use the ideal gas law to calculate the volume of O 2 at STP. The ideal gas law is: P V = n RT At STP, the pressure P = 1 atm, and the temperature T = 273.15 K. The ideal gas constant R = 0.0821 L atm / (mol K). We can rearrange the ideal gas law to solve for the volume V: V = P n RT V O 2 = 1 a t m ( 0.0829 m o l ) ( 0.0821 L a t m / ( m o l K )) ( 273.15 K ) = 1.859 L
Convert volume to milliliters Finally, we need to convert the volume from liters to milliliters. Since 1 L = 1000 mL: V O 2 ( m L ) = V O 2 ( L ) ∗ 1000 V O 2 ( m L ) = 1.859 L ∗ 1000 m L / L = 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. The closest answer choice is 1960 mL.
Examples
This calculation is crucial in various fields, such as designing airbags in cars. The rapid reaction between chemicals produces gas to inflate the airbag, protecting the occupant during a collision. Knowing the precise volume of gas produced from a given amount of reactant ensures the airbag inflates to the correct size and pressure, providing optimal safety. Similarly, in industrial processes involving gas-producing reactions, accurate volume calculations are essential for safety and efficiency.
