The volume of hydrogen gas formed when 1.5g of aluminum reacts with aq NaOH at 27 degrees Celcius is 1.378 L.
Number of moles of hydrogen formed
The number of moles of hydrogen formed is determined from the given equation as follows;
Al + NaOH + H2O -------------> H2 + NaAlO2
from the equation above;
1 mole of aluminum ------> 1 mole of hydrogen
1.5 g/27g = 0.056 mole of hydrogen
Volume of hydrogen formed
The volume of hydrogen formed is determined as follows;
at STP, 1 mole of hydrogen gas is equal to 22.4 L of hydrogen gas.
0.056 mole = ?
= 1.254 L
at STP, temperature = 273 K, and pressure = 1 atm
27⁰C = 300 K
V₁/T₁ = V₂/T₂
V₂ = (V₁/T₁) x T₂
V₂ = (1.254/273) x 300
V₂ = 1.378 L
Thus, the volume of hydrogen gas formed when 1.5g of aluminum reacts with aq NaOH at 27 degrees Celcius is 1.378 L.
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