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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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3 years ago

Show the equation how many grams of Mg(OH)2 would be produced from 4 mol KOH?

Sindrei [870]

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3 years ago

On the basis of the information above, what is the approximate percent ionization of HNO2 in a 1.0 M HNO2 (aq) solution?

enot [183]

Answer:

The answer is "2%"

Explanation:

Equation:

$HNO_2\ (aq) \leftrightharpoons H^{+} \ (aq) + NO_2^{-}\ (aq) \\\\\ K_a = 4.0\times \ 10^{-4}$

$H^{+}=?$

Formula:

$Ka = \frac{[H^{+}][NO_2^{-}]}{[HNO_2]}$

Let

$[H^{+}] = [NO_2^{-}] = x$ at equilibrium

$x^2 = (4.0\times 10^{-4})\times 1.0\\\\x = ((4.0\times 10^{-4})\times 1.0)^{0.5} = 2.0 \times 10^{-2} \ M\\\\$

therefore,

$[H^{+}] = 2.0\times 10^{-2} \ M = 0.02 \ M$

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