Part A: How many liters of H2 gas can be produced at 0 C and 1.00 atm (STP) from

ID: 998802 • Letter: P

Question

Part A:

How many liters of H2 gas can be produced at 0 C and 1.00 atm (STP) from 40.0 g of Zn ?
Zn(s)+2HCl(aq)ZnCl2(aq)+H2(g)

Part B:

A steel cylinder with a volume of 15.0 L is filled with 64.2 g of nitrogren gas at 25C .

What is the pressure, in atmospheres, of the N2 gas in the cylinder?

Part C:

A gas mixture containing oxygen, nitrogen, and neon exerts a total pressure of 1.16 atm .

If helium added to the mixture increases the pressure to 1.74 atm , what is the partial pressure (atm) of the helium?

1. 40 g of Zn is equal to 0.61 mol of Zn (n=m/M where M= 65.3 g/mol)

Because Zn and H2 is in a 1:1 ratio in equation, there is also 0.61 mol H2 produced.

Use PV= nRT to find V

V= nRT/P = 0.61mol*0.082*273K / 1atm= 13.7 L

2. Use PV=nrT to find P

P= nRT/V = 2.3 mol*0.082*(25+273)/15= 3.75 atm

mole of N2 is n=m/M=64.2/28= 2.3 mol

3. amount of helium added is 1.74-1.16 = 0.58 atm

partial pressure= 0.58/ 1.74= 0.33 atm