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0.9g of Al reacts with dil.HCl to give H2.The volume of H2 gas evolved at 273 degree Celsius and 1 atm pressure will be 1)1.12 L 2)2.24 L 3)5.6 L 4)0.112 L?
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0.9g of Al reacts with dil.HCl to give H2.The volume of H2 gas evolved...
To determine the volume of hydrogen gas evolved when 0.9g of Al reacts with dilute hydrochloric acid (HCl) at 273 degrees Celsius and 1 atm pressure, we need to use the ideal gas law and stoichiometry.

1. Calculate the moles of Al:
The molar mass of Al is 26.98 g/mol.
moles of Al = mass of Al / molar mass of Al
moles of Al = 0.9g / 26.98 g/mol
moles of Al = 0.0334 mol

2. Write and balance the chemical equation for the reaction:
2Al + 6HCl → 2AlCl3 + 3H2

From the balanced equation, we can see that 2 moles of Al react to produce 3 moles of H2.

3. Calculate the moles of H2:
moles of H2 = (moles of Al) x (3 moles of H2 / 2 moles of Al)
moles of H2 = 0.0334 mol x (3/2)
moles of H2 = 0.0501 mol

4. Use the ideal gas law to calculate the volume of H2 gas at STP:
The ideal gas law equation is PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature in Kelvin.

Since the pressure is given as 1 atm and the temperature is given as 273 degrees Celsius, we need to convert the temperature to Kelvin:
T(K) = 273 + 273 = 546 K

Now we can rearrange the ideal gas law equation to solve for volume:
V = (nRT) / P
V = (0.0501 mol x 0.0821 L·atm/(mol·K) x 546 K) / 1 atm
V = 2.24 L

Therefore, the volume of H2 gas evolved at 273 degrees Celsius and 1 atm pressure is 2.24 L (Option 2).
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0.9g of Al reacts with dil.HCl to give H2.The volume of H2 gas evolved...
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0.9g of Al reacts with dil.HCl to give H2.The volume of H2 gas evolved at 273 degree Celsius and 1 atm pressure will be 1)1.12 L 2)2.24 L 3)5.6 L 4)0.112 L?
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