To calculate the volume of chlorine gas that reacted with 0.84 grams of aluminum, we need to use the stoichiometry of the reaction between aluminum and chlorine.
The balanced chemical equation for the reaction is:
2Al + 3Cl2 -> 2AlCl3
This means that 2 moles of aluminum react with 3 moles of chlorine gas to produce 2 moles of aluminum chloride.
First, we need to find the number of moles of aluminum in 0.84 grams. The molar mass of aluminum is 26.98 g/mol, so:
0.84 g Al / 26.98 g/mol Al = 0.031 moles Al
According to the stoichiometry of the balanced equation, 2 moles of aluminum react with 3 moles of Cl2. Therefore, we can use a proportion to calculate the number of moles of Cl2 that react with 0.031 moles of Al:
2 moles Al : 3 moles Cl2 = 0.031 moles Al : x moles Cl2
x = (0.031 moles Al)(3 moles Cl2 / 2 moles Al) = 0.047 moles Cl2
Finally, we can use the ideal gas law to calculate the volume of chlorine gas that corresponds to 0.047 moles at standard temperature and pressure (STP).
PV = nRT
where P = 1 atm (STP pressure), V is the volume of chlorine gas, n = 0.047 moles, R = 0.082 L·atm/mol·K (gas constant), and T = 273 K (STP temperature).
Solving for V:
V = nRT / P = (0.047 mol)(0.082 L·atm/mol·K)(273 K) / (1 atm) ≈ 1.1 L
Therefore, the volume of chlorine gas that reacted with 0.84 grams of aluminum is approximately 1.1 liters.
0.84grams or aluminum reacted completely with chlorine gas . calculate the volume of chlorine
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