If 0.836 g of solid Al reacts stoichiometrically according to the balanced equation in a reaction solution with a total volume of 1160 mL, what mass (g) of gaseous H2 is produced?
6 HClO4(aq) + 2 Al(s) → 3 H2(g) + 2 Al(ClO4)3(aq)
0.836g/molecular weight Al=mole of Al
mole of Al x3/2=mole of H2
mole of H2 x molecular weight H2=gram H2
That looks ok to me.
To find the mass of gaseous H2 produced, we need to follow these steps:
1. Calculate the number of moles of Al: Divide the given mass of Al (0.836 g) by the molar mass of Al. The molar mass of Al is 26.98 g/mol.
0.836 g / 26.98 g/mol = 0.031 mol Al
2. Apply the stoichiometric ratio between Al and H2 from the balanced equation to determine the number of moles of H2 produced. From the balanced equation, we see that 2 moles of Al react to produce 3 moles of H2.
0.031 mol Al x (3 mol H2 / 2 mol Al) = 0.0465 mol H2
3. Calculate the mass of H2 produced by multiplying the number of moles of H2 by its molar mass. The molar mass of H2 is 2.02 g/mol.
0.0465 mol H2 x 2.02 g/mol = 0.093 g H2
Therefore, the mass of gaseous H2 produced is 0.093 g.