Question

What volume (in liters) of carbon monoxide gas, measured at a temperature of 212 K and a pressure of 676 mm Hg, is required to synthesize 19.3 g of methanol. How many liters of oxygen (at STP) are required to form 12.5 g of H2O ? Show your work

Answer 1

For 1: The volume of carbon monoxide required to produce given amount of methanol is 11.9 L.

For 2: The volume of oxygen required to form given amount of water is 7.77 L.

Step-by-step explanation:

• For 1:

To calculate the number of moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` ......(1)

For methanol:

Given mass of methanol = 19.3 g

Molar mass of methanol = 32 g/mol

Putting values in above equation, we get:

`\text{Moles of methanol}=(19.3g)/(32g/mol)=0.603mol`

The chemical reaction of formation of methanol from carbon monoxide follows:

`CO+2H_2\rightarrow CH_3OH`

By Stoichiometry of the reaction:

1 mole of methanol is produced from 1 mole of carbon monoxide.

so, 0.603 moles of methanol will be produced from =
`(1)/(1)* 0.603=0.603mol` of carbon monoxide.

To calculate the volume of carbon monoxide, we use the equation given by ideal gas:

PV = nRT

where,

P = Pressure of carbon monoxide = 676 mmHg

V = Volume of carbon monoxide = ? L

n = Number of moles of carbon monoxide = 0.603 mol

R = Gas constant =
`62.3637\text{ L.mmHg }mol^(-1)K^(-1)`

T = Temperature of carbon monoxide = 212 K

Putting values in above equation, we get:

`676mmHg* V=(19.3)/(32g/mol)* 62.3637\text{ L.mmHg }mol^(-1)K^(-1)* 212K\\\\V=11.79L`

Hence, the volume of carbon monoxide required to produce given amount of methanol is 11.9 L.

• For 2:

Calculating the moles of water by using equation 1, we get:

Given mass of water = 12.5 g

Molar mass of water = 18 g/mol

Putting values in equation 1, we get:

`\text{Moles of water}=(12.5g)/(18g/mol)=0.694mol`

The chemical reaction of formation of water from oxygen and hydrogen follows:

`O_2+2H_2\rightarrow 2H_2O`

By Stoichiometry of the reaction:

2 moles of water is produced from 1 mole of oxygen gas.

so, 0.694 moles of water will be produced from =
`(1)/(2)* 0.694=0.347mol` of oxygen gas.

At STP:

1 mole of a gas occupies 22.4 L of volume.

So, 0.347 moles of oxygen gas will occupy =
`(22.4L)/(1mol)* 0.347mol=7.77L`

Hence, the volume of oxygen required to form given amount of water is 7.77 L.