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mario62 [17]
3 years ago
8

How many moles of iron(III) hydroxide precipitate will form when 2.7 moles of aqueous sodium hydroxide reacts completely with ex

cess iron(III) nitrate solution according to the following reaction? (record your answer to 1 decimal place)
Chemistry
1 answer:
aalyn [17]3 years ago
7 0

Answer:

0.9 mole of Fe(OH)3.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

Fe(NO3)3 + 3NaOH —> Fe(OH)3 + 3NaNO3

Now, we can determine the moles of iron (III) hydroxide formed from the reaction as follow:

From the balanced equation above,

3 moles of NaOH reacted to produce 1 mole of Fe(OH)3.

Therefore, 2.7 moles of NaOH will react to produce = 2.7/3 = 0.9 mole of Fe(OH)3.

Therefore, 0.9 mole of Fe(OH)3 is produced from the reaction.

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In the spring of 1984, concern arose over the presence of ethylene dibromide, or EDB, in grains and cereals. EDB has the molecul
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869.6 × 10¹⁴ molecules of EDB

Explanation:

We have 1.9 lb of flour with a EDB concentration of 31.5 ppb.

We need to transform lb in grams.

1 lb = 453.6 grams

1.9 lb = (1.9 × 453.6) / 1 = 861.8 grams

Now we determine the number of molecules of EDB in the sample by devise the following reasoning:

if we have        31.5 × 10⁻⁹ g of EDB in 1 g of sample

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Taking in account that 1 mole of any substance contains 6.022 × 10²³ (Avogadro’s number) molecules we devise the following reasoning:

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