Answer:
1.82g of Zn remain
Explanation:
Zinc, Zn, reacts with hydrochloric acid, HCl, as follows:
Zn + 2HCl → ZnCl2 + H2
Where 1 mole of Zn reacts with 2 moles of HCl
To solve this question we must find the moles of each reactant. With the moles of HCl we can find the moles of zinc needed for a complete reaction and the mass of zinc that remains:
Moles Zn -Molar mass: 65.32g/mol-
6.72g * (1mol / 65.38g) = 0.1028 moles Zn
Moles HCl:
100.0mL = 0.1000L * (1.50mol / L) = 0.150moles HCl
The moles of Zn that react with 0.150 moles HCl are:
0.150 mol HCl * (1mol Zn / 2mol HCl) = 0.075 moles of Zn are required
The moles that remain are:
0.1028 moles Zn - 0.075 moles of Zn = 0.0278 moles Zn
The mass is:
0.0278 moles Zn * (65.38g / mol) =
1.82g of Zn remain
