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madreJ [45]
3 years ago
10

Answer the following question: Ethanol, C2H5OH, is considered clean fuel because it burns in oxygen to produce carbon dioxide an

d water with few trace pollutants. If 150.0 g of H2O are produced during the combustion of ethanol, how many grams of ethanol were present at the beginning of the reaction?
Chemistry
1 answer:
Delvig [45]3 years ago
7 0

Answer:

127.88 grams of ethanol were present at the beginning of the reaction

Explanation:

Firstly, let's make the combustion reaction:

C₂H₅OH  +  3O₂  →  2CO₂  +  3H₂O

By 1 mol of ethanol, you can make 3 mole of water.

Mole of water = Water mass / Molar mass

150g / 18g/m = 8.3 mole

3 mole of water came from 1 mol of ethanol

8.3 mole came from (8.3 .1)/3 = 2.78 mole of ethanol

Molar mass ethanol = 46 g/m

Mole . molar mass = mass

2.78 m .  46g/m = 127.88 g

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How many total atoms are in 0.290 g of P2O5?
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There are 8.61 × 10²⁰ atoms in 0.290 g P₂O₅.

Step 1. Convert <em>grams of P₂O₅ to moles of P₂O₅</em>.

\text{Moles of P}_{2}\text{O}_{5} = \text{0.290 g } \text{P}_{2}\text{O}_{5} \times \frac{\text{1 mol }\text{P}_{2}\text{O}_{5}}{\text{141.94 g }\text{P}_{2}\text{O}_{5}} = \text{0.002 043 mol } \text{P}_{2}\text{O}_{5} \\

Step 2. Convert <em>moles of P₂O₅ to molecules of P₂O₅</em>.

\text{Molecules of } \text{P}_{2}\text{O}_{5} = \text{0.002 043 mol } \text{P}_{2}\text{O}_{5} \times \frac{6.022 \times 10^{23}\text{ molecules }\text{P}_{2}\text{O}_{5}}{\text{1 mol } \text{P}_{2}\text{O}_{5}}\\

= 1.23 \times10^{21}\text{ molecules } \text{P}_{2}\text{O}_{5}\\

Step 3. Convert <em>molecules of P₂O₅ to atoms</em>.

There are seven atoms in 1 mol P₂O₅.

∴ \text{Total atoms} = 1.23\times 10^{21 }\text{ molecules }\text{P}_{2}\text{O}_{5} \times\frac{\text{7 atoms}}{\text{1 molecule }\text{P}_{2}\text{O}_{5}} = 8.61 \times 10^{21}\text{ atoms}\\

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