2.7g of aluminum metal is placed into 200g of solution hydrochloric acid to form aluminum
chloride and hydrogen gas.
a. Write the chemical equation.
b. Find the mass of the obtained aluminum
chloride.
C. Find the percent by mass of the acidic solution.
d. Find the volume of the obtained hydrogen gas
at STP.

a.

2Al + 6HCl--> 2AlCl3 + 3H2

b.

take the atomic mass of Al =27.0, Cl = 35.5, and H= 1.0,

no. of moles = .mass / molar mass

no. of moles of Al placed = 2.7 / 27.0

=0.1 mol

Assuming Al is limiting (because the concentration nor volume of HCl solution is given).

From the equation, the mole ratio of Al : AlCl3 = 2:2 = 1: 1,

meaning for each 1 mole of Al reacted, it gives out 1 mole of AlCl3. This also means the no. of moles of Al reacted equals to the no. of moles of AlCl3 produced, which is 0.1 mole.

mass = no. of moles x molar mass

hence,

mass of AlCl3 produced = 0.1 / (27 + 35.5x3)

= 0.00075 g

c. since this question is unclear, and the concentration nor volume of HCl solution is given, part c cannot be answered. Please let me know if you have any more info regarding it! thanks :)

d. From the equation, the mole ratio of Al:H2 = 2:3, meaning every 2 moles of Al reacted, gives out 3 moles of hydrogen.

using this ratio,  no. of moles of H2 produced will = 0.1 x 3 / 2

=0.15 mol

At STP, the molar volume of gas is 22.4 dm3 mol-1,

volume = no. of moles x molar volume

hence volume of H2 obtained = 0.15 x 22.4

= 3.36 dm3