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2 years ago

When 84. 8 g of iron (III) oxide reacts with an excess of carbon

Chemistry

1 answer:

KengaRu [80]2 years ago

7 0

Answer:

Use the balanced equation to determine the theoretical yield. Focus on the chemicals whose quantities you are given or are asked to find. CO is not important as it is excess and will make sure that all of the Fe2O3 will be changed to iron. Remember the quotients (numbers in front of formula) are the number of moles. You are given the mass of Fe2O3 in grams so you will need to convert 1 mole of Fe2O3 into grams (molar mass which, using simple atomic masses = 56x2 + 16x3 = 160g/mol but you use atomic masses that your course requires

Fe2O3(s) + 3CO(g)--> 2Fe(s) + 3CO2(g)

1 mol 2 mol

160g 112g

84.8g (112/160) x 84.8

So now you have the theoretical yield of Fe2O3 and the experimental (actual) yield = 54.3g)

Percentage Yield = (actual yield/theoretical yield) x 100

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Answer:

$Al_2(SO_4)_3+3Mg(NO_3)_2\rightarrow 2Al(NO_3)_3+3MgSO_4$

Explanation:

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In this case, for the given reactants side, we infer this is a double replacement reaction because all the cations and anions are switched around as a result of the chemical change, we infer that the products side include aluminum with nitrate and magnesium with sulfate as shown below:

$Al_2(SO_4)_3+Mg(NO_3)_2\rightarrow Al(NO_3)_3+MgSO_4$

However, we need to balance since unequal number of atoms are present at both sides, thus, we do that as shown below:

$Al_2(SO_4)_3+3Mg(NO_3)_2\rightarrow 2Al(NO_3)_3+3MgSO_4$

Thus, we make 6 Al atoms, 3 S atoms, 3 Mg atoms and 30 O atoms on each side in agreement with the law of conservation of mass.

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$\text{engines, since its reaction with oxygen produces only nitrogen and water vapor,}$

$\text{and in the liquid form it is easily transported. An industrial chemist studying this}$

$\text{reaction fills a} \ \mathbf{100 \ L }\ \text{tank with} \ \mathbf{8.6 \ mol} \ \text{of ammonia gas and} \ \mathbf{28 \ mol} \ \ \text{of oxygen gas, }$

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Answer:

Explanation:

From the correct question above:

The reaction can be represented as:

$\mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }$

From the above reaction; the ICE table can be represented as:

$\mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }$

I (mol/L) 0.086 0.28 0 0

C -4x -3x +2x +6x

E 0.086 - 4x 0.28 - 3x +2x +6x

At equilibrium;

The water vapor = $\dfrac{2.6 \ mol}{100 \ L} = 6x$

$x = \dfrac{2.6}{100} \times \dfrac{1}{6}$

$x = 0.00433$

$\text{equilibrium constant} ({k_c}) = \dfrac{ [N_2]^2 [H_2O]^6 }{ [[NH_3]^4] [O_2]^3 }$

$\implies \dfrac{(2x)^2 (6x)^6}{(0.086-4x)^4\times (0.28-3x)^3} \\ \\$

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