What percentage of the mass of 1 mole of CH3Cl is derived from carbon?

Chemistry

1 answer:

steposvetlana [31]3 years ago

3 0

Answer:

Element Symbol Atomic weight Atoms Mass percent

Carbon C 12.0107 1 23.7894

Hydrogen H 1.00794 3 5.9892

Chlorine Cl 35.453 1 70.2213

Explanation:

You might be interested in

Blast furnaces extra pure iron from the Iron(IIl)oxide in iron ore in a two step sequence. In the first step, carbon and oxygen

OLga [1]

Answer:

5.9 kg

Explanation:

We must work backwards from the second step to work out the mass of oxygen.

1. Second step

Mᵣ: 55.84

Fe₂O₃ + 3CO ⟶ 2Fe + 3CO₂

m/kg: 7.0

(a) Moles of Fe

$\text{Moles of FeO} = \text{7000 g Fe} \times \dfrac{\text{1 mol Fe}}{\text{55.84 g Fe}} = \text{125 mol Fe}$

(b) Moles of CO

$\text{Moles of CO} = \text{125 mol Fe} \times \dfrac{\text{3 mol CO}}{\text{2 mol Fe}} = \text{188 mol CO}$

However, this is the theoretical yield.

The actual yield is 72. %.

We need more CO and Fe₂O₃ to get the theoretical yield of Fe.

(c) Percent yield

$\begin{array}{rcl}\text{Percent yield} &=& \dfrac{\text{ actual yield}}{\text{ theoretical yield}} \times 100 \, \%\\\\ 72. \, \% & = & \dfrac{\text{188 mol}}{\text{actual yield}} \times 100 \,\%\\\\0.72 &= &\dfrac{\text{188 mol}}{\text{actual yield}}\\\\\text{Actual yield} & = & \dfrac{\text{188 mol}}{0.72}\\& = & \textbf{261 mol}\\\\\end{array}$

We must use 261 mol of CO to get 7.0 kg of Fe.

2. First step

Mᵣ: 32.00

2C + O₂ ⟶ 2CO

n/mol: 261

(a) Moles of O₂

$\text{Moles of O}_{2} = \text{261 mol CO} \times \dfrac{\text{1 mol O}_{2}}{\text{2 mol CO}} = \text{131 mol O}_{2}$

(b) Mass of O₂

$\text{Mass of O}_{2}= \text{131 mol O }_{2} \times \dfrac{\text{32.00 g O}_{2}}{\text{1 mol O}_{2}} = \text{4180 g O}_{2}$

However, this is the theoretical yield.

The actual yield is 71. %.

We need more C and O₂ to get the theoretical yield of CO.

(c) Percent yield

$\begin{array}{rcl}71. \, \% & = & \dfrac{\text{188 mol}}{\text{actual yield}} \times 100 \,\%\\\\0.71 &= &\dfrac{\text{4180 g}}{\text{actual yield}}\\\\\text{Actual yield} & = & \dfrac{\text{4180 g}}{0.71}\\\\& = & \text{5900 g}\\& = & \textbf{5.9 kg}\\\end{array}$