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sashaice [31]
3 years ago
14

Consider the graph below. Which phase change is most likely taking place in this graph? a gas to a liquid a liquid to a solid a

gas to a solid a solid to a liquid

Chemistry
1 answer:
luda_lava [24]3 years ago
4 0
A solid to a liquid
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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
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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

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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}

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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}

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