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18 November, 22:21

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

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  1. 19 November, 02:01
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    0.0425 g

    Explanation:

    We are given;

    Temperature, T for the day as 22°C

    but, K = °C + 273.15

    Thus, T = 295.15 K

    Barometric pressure, P = 742 mmHg Volume of the gas, V = 43.5 mL or 0.0435 L

    We are required to calculate the mass of magnesium that was used in the reaction.

    Step 1: Number of moles of the gas produced

    Using the ideal gas equation, we can determine the number of moles

    PV = nRT, where R is the ideal gas constant, 62.3637 L·mmHg/mol·K

    Rearranging the formula;

    n = PV : RT

    = (742 mmHg * 0.0435 L) : (62.3637 * 295.15K)

    = 0.00175 moles

    Step 2: Moles of magnesium that reacted

    The equation fro the reaction is;

    Mg + 2HCl → MgCl₂ + H₂

    Thus, 1 mole of Mg reacts to produce 1 mole of hydrogen gas

    Therefore, moles of Mg = Moles of Hydrogen gas

    = 0.00175 moles

    Step 3: Mass of magnesium that reacted

    We know that, Moles = Mass : Molar mass

    Thus, Mass = Moles * Molar mass

    Molar mass of Mg = 24.305 g/mol

    Hence;

    Mass of Mg = 0.00175 moles * 24.305 g/mol

    = 0.0425 g

    Thus, 0.0425 g of magnesium reacted
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