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Determine the volume of gas that would be produced when 9 cm of magnesium ribbon reacts with the acid at rtp. 1 mole of gas occupies 24 dm3 at rtp

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Final answer:

To determine the volume of gas produced when 9 cm of magnesium ribbon reacts with the acid at RTP, calculate the number of moles of magnesium, determine the number of moles of gas produced using stoichiometry, and convert the moles of gas to volume using the given information.

Step-by-step explanation:

To determine the volume of gas produced when 9 cm of magnesium ribbon reacts with the acid at RTP, we first need to calculate the number of moles of magnesium. We can then use the stoichiometry of the reaction to determine the number of moles of gas produced, which can be converted to volume using the given information that 1 mole of gas occupies 24 dm³ at RTP.

Step 1: Calculate the number of moles of magnesium

Molar mass of Mg = 24.31 g/mol

Number of moles = mass / molar mass

= 9 g / 24.31 g/mol

= 0.37 mol

Step 2: Determine the number of moles of gas produced

From the balanced chemical equation, we know that 1 mole of magnesium produces 1 mole of gas.

So, the number of moles of gas produced = 0.37 mol

Step 3: Convert the number of moles of gas produced to volume

1 mole of gas = 24 dm³

So, the volume of gas produced = 0.37 mol × 24 dm³/mol

= 8.88 dm³

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