20.6k views
4 votes
Use the following RXN

Zn+2HCI=ZnCI2 + H2

a) What mass of zinc is required to form 1 L of the gas at STP

b) What volume of the gas at 22 C and o.976 atm can be prepared from 1 g Zn?

Please answer this question as soon as you can.

User Brablc
by
5.2k points

1 Answer

3 votes

Answer:

a) 2.94 g of zinc

b) volume of gas = 0.379 L

Step-by-step explanation:

Part 1:

Data Given:

volume of gas (H₂) = 1 L

mass of Zn = ?

Reaction Given:

Zn+2HCI -------> ZnCI₂ + H₂

Solution:

To find mass of zinc first find number of mole of 1 L gas (H₂)

formula used

no. of mole = volume of gas/ molar volume at STP

As we Know

molar volume at STP = 22.414 mol/L

Put value in above formula

no. of mole = 1 L / 22.414 (mol/L)

no. of mole = 0.045 mol

So

1 L of gas (H₂) = 0.045 mol

Now Look at the reaction

Zn + 2 HCI -------> ZnCI₂ + H₂

1 mol 1 mol

As 1 mole of zinc form 1 mole of Hthen how many moles will require to form 0.045 mole of gas (H₂).

Apply unity formula

1 mole of zinc ≅ 1 mole of H₂

X moles of zinc ≅ 0.045 mole of H₂

Do cross multiplication

moles of zinc = 1 mole x 0.045 mole / 1 mole

moles of zinc = 0.045 mole

To find mass of Zinc, convert no moles to mass

Formula Used

mass in grams = no. of moles x molar mass . . . . . . . .(1)

molar mass of zinc = 65.4 g/mol

Put values in equation 1

mass in grams = 0.045 mole x 65.4 g/mol

mass in grams = 2.94 g

So,

2.94 g of zinc is required to form 1 L of the gas at STP

____________

Part 2.

Data given:

Temperature (T) = 22 °C

Pressure (P) = 0.976 atm

convert °C to kelvin

T = °C + 273

T = 22 °C + 273

T = 295 K

volume of gas = ?

Mass of zinc = 1 g

Solution:

To find volume formula used

PV= nRT

Rearrange the equation for Volume

V = nRT / P . . . . . . . . . (1)

where

P = pressure

V = Volume

T= Temperature

n = Number of moles

R = ideal gas constant = 0.08206 L.atm / mol. K

For this we have to know the mole of the znic

The formula will be used

no. of moles = mass in grams / molar mass . . . . . . (2)

Molar mass of Zn = 65.4 g/mol

Put values in equation 2

no. of moles = 1 g / 65.4 g/mol

no. of moles = 0.0153 mol

Now put the value in formula (2) to calculate volume of gas

V = 0.0153 mol x 295 K x 0.08206 (L.atm/mol. K) / 0.976 atm

V = 0.3699 (L.atm) /0.976 atm

V = 0.379 L

So,

The volume of gas will be 0.379 L

User Urvish
by
5.6k points