Question

2.70g of Zn (s) reacts with 50.0 mL of 1.00 M HCl solution to produce hydrogen gas according to the reaction. (R = 0.08206 L·atm·K-1·mol-1) Zn(s) + 2HCl (aq) = ZnCl2 (aq) + H2(g) a) calculate mole of Zn used in this reaction b) calculate mole of HCl used in this reaction c) calculate which reactant is a limiting reactant d) calculate mole of hydrogen gas formed e) calculate volume of H2 gas at STP in Liters that will be produced during the reaction

Answer 1

a. 0.0413 moles Zn

b. 0.0500 moles HCl

c. HCl is the limiting reactant

d. 0.0250 moles H₂

e. V = 0.56L

Step-by-step explanation:

The reaction of Zn(s) with HCl is:

Zn(s) + 2HCl (aq) → ZnCl₂ (aq) + H₂(g)

Where 1 mole of Zn reacts with 2 moles of HCl.

a) To convert mass in grams to moles of a substance you need to use molar mass (Molar mass Zn: 65.38g/mol), thus:

2.70g Zn × (1mol / 65.38g) = 0.0413moles of Zn

b. Now, when you have a solution in molarity (Moles / L), you can know the moles of a volume of solution, thus:

Moles HCl:

50.0mL = 0.0500L × (1.00mol / L) = 0.0500 moles HCl

c. The limiting reactant is founded by using the chemical reaction as follows:

For a complete reaction of 0.0500 moles HCl you need:

0.0500 moles HCl × (1 mole Zn / 2 moles HCl) = 0.0250 moles Zn

As you have 0.0413 moles of Zn, and you need just 0.0250 moles for the complete reaction, Zn is the exces reactant and HCl is the limiting reactant

d.As HCl is limiting reactant and 2 moles of HCl react with 1 mole of H₂, moles of hydrogen formed are:

0.0500 moles HCl × (1 mole H₂ / 2 moles HCl) = 0.0250 moles H₂

e. Using PV = nRT, you can find volume of gas, thus:

PV = nRT

V = nRT / P

Where P is pressure 1atm at STP, n are moles, R is gas consant 0.08206Latm/molK and T is absolute temperature 273.15K at STP.

V = 0.0250molesₓ0.082atmL/molKₓ273.15K / 1atm

V = 0.56L