Question

10^21 molecules are removed from 200mg of CO2.The moles of CO2 left are?

Answer 1

$\checkmark$ Given:

10²¹ molecules are removed from 200 mg of CO₂

$\checkmark$ To Find:

The moles of CO₂ left

$\checkmark$ Solution:

Molar Mass of CO₂ = 44 g/mol

Given Mass of CO₂ = 200 mg = 0.2 g

Calculating No. of Mole

$\implies \rm Mole=\dfrac{W}{GMW}$

Here,

• W = Given Mass

• GMW = Molar Mass

Hence,

• W = 0.2 g

• GMW = 44 g/mol

By Substituting the values,

We get,

$\rm \implies M=\dfrac{0.2}{44} \ moles$

$\rm \implies M=0.00454 \ moles$

$\rm \implies M=4.55 \times 10^{-3} \ moles$

$\rm \implies Initial = 4.55 \times 10^{-3} \ moles$

Given that,

10²¹ molecules are removed from 200 mg of CO₂

Hence,

No. of Molecules of CO₂ removed = 10²¹

We Know that,

$\rm 6.023 \times 10^{23} \ molecules \ of \ CO_{2}=1 \ mole$

Hence,

$\rm 10^{21} \ molecules \ of \ CO_{2}= "x" \ moles$

By Cross-Multiplication,

We get,

$\rm \implies x=\dfrac{10^{21}}{6.023 \times 10^{23}} \ moles$

$\rm \implies x=0.1660 \ moles$

$\rm \implies x=1.66 \times 10^{-3} \ moles$

$\rm \implies Final = 1.66 \times 10^{-3} \ moles$

According to the Question,

We are asked to find The moles of CO₂ left

Hence,

Moles of CO₂ left = Initial - Final

Moles of CO₂ left = 4.55 x 10⁻³ - 1.66 x 10⁻³

Moles of CO₂ left = (4.55 - 1.66) x 10⁻³

Moles of CO₂ left = 2.89 x 10⁻³

Answer 2

Answer:

$\checkmark$ Given:

10²¹ molecules are removed from 200 mg of CO₂

$\checkmark$ To Find:

The moles of CO₂ left

$\checkmark$ Solution:

Molar Mass of CO₂ = 44 g/mol

Given Mass of CO₂ = 200 mg = 0.2 g

Calculating No. of Mole

$\implies \rm Mole=\dfrac{W}{GMW}$

Here,

W = Given Mass

GMW = Molar Mass

Hence,

W = 0.2 g

GMW = 44 g/mol

By Substituting the values,

We get,

$\rm \implies M=\dfrac{0.2}{44} \ moles$

$\rm \implies M=0.00454 \ moles$

$\rm \implies M=4.55 \times 10^{-3} \ moles$

$\rm \implies Initial = 4.55 \times 10^{-3} \ moles$

Given that,

10²¹ molecules are removed from 200 mg of CO₂

Hence,

No. of Molecules of CO₂ removed = 10²¹

We Know that,

$\rm 6.023 \times 10^{23} \ molecules \ of \ CO_{2}=1 \ mole$

Hence,

$\rm 10^{21} \ molecules \ of \ CO_{2}= "x" \ moles$

By Cross-Multiplication,

We get,

$\rm \implies x=\dfrac{10^{21}}{6.023 \times 10^{23}} \ moles$

$\rm \implies x=0.1660 \ moles$

$\rm \implies x=1.66 \times 10^{-3} \ moles$

$\rm \implies Final = 1.66 \times 10^{-3} \ moles$

According to the Question,

We are asked to find The moles of CO₂ left

Hence,

Moles of CO₂ left = Initial - Final

Moles of CO₂ left = 4.55 x 10⁻³ - 1.66 x 10⁻³

Moles of CO₂ left = (4.55 - 1.66) x 10⁻³

Moles of CO₂ left = 2.89 x 10⁻³