Aluminum weighing 0.81 g was introduced into a solution containing 4.9 g of sulfuric acid. What volume of hydrogen will be released in the reaction?
2Al + 3H2SO4 = Al2 (SO4) 3 + 3H2
It can be seen from the equation that 2 aluminum molecules with a mass of 0.81 g and 3 molecules of sulfuric acid with a given mass of 4.9 g enter into the reaction.
Let’s find the number of moles of these substances.
n = m / M
M (Al) = 27 • 2 = 54 g / mol
M (H2SO4) = 98 • 3 = 294 g / mol
n (Al) = 0.81 / 54 = 0.015 mol (deficiency)
n (H2SO4) = 4.9 / 294 = 1.017 mol (excess)
The molar volume of hydrogen under normal conditions is 22.4 l / mol, respectively, from the reaction 3 • 22.4 = 67.2 l / mol
Let’s find the volume of hydrogen in terms of the amount of aluminum:
V (H2) = 0.81 • 67.2 / 54 = 1.008 l
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