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A student performed an experiment in the same manner as part B (magnesium+HCL). The temperature for the day was 22Ā°C in the barometric pressure was 742MMHG. 43.5 mL of gas was collected how many grams of magnesium mustā€™ve been used in the reaction?

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Answer:

Mass = Ā 0.048 g

Explanation:

Given data:

Temperature =22 Ā°C (273+ 22 = 296K)

Pressure = 742 mmHg (742/760 = 0.98 atm)

Volume = 43.5 mL ( 0.044 L)

Mass of magnesium = ?

Solution:

PV = nRT

n = PV/RT

n = Ā 0.98 atm Ɨ 0.044 L / 0.0821 atm. L. molā»Ā¹. Kā»Ā¹ Ɨ 296K)

n = 0.043 atm. L / 24.3 atm. L. molā»Ā¹

n = 0.002 mol

Mass of Mg:

Mass = number of moles Ɨ Ā molar mass

Mass = Ā 0.002 mol Ɨ 24 g/mol

Mass = Ā 0.048 g

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