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blagie [28]
2 years ago
9

An unknown amount of Al203 decomposed producing 215 g of solid aluminum. 2Al2O3=4Al+3O2 How many grams of oxygen gas should be p

roduced
Chemistry
1 answer:
natulia [17]2 years ago
6 0

Answer:

191.11 grams of oxygen gas should be produced.

Explanation:

The balanced reaction is:

2 Al₂O₃ → 4 Al + 3 O₂

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • Al₂O₃: 2 moles
  • Al: 4 moles
  • O₂: 3 moles

Being the molar mass of each compound:

  • Al₂O₃: 102 g/mole
  • Al: 27 g/mole
  • O₂: 32 g/mole

By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • Al₂O₃: 2 moles* 102 g/mole= 204 grams
  • Al: 4 moles* 27 g/mole= 108 grams
  • O₂: 3 moles* 32 g/mole= 96 grams

Then you can apply the following rule of three: if by stoichiometry 108 grams of aluminum are produced along with 96 grams of oxygen, 215 grams of aluminum are produced along with how much mass of oxygen?

mass of oxygen=\frac{215 grams of aluminum*96 grams of oxygen}{108grams of aluminum}

mass of oxygen= 191.11 grams

<u><em>191.11 grams of oxygen gas should be produced.</em></u>

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SnO2 + 2 H2 ——&gt; Sn + 2 H2O
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Answer:

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Explanation:

Given parameters:

Number of molecules of water = 1.2 x 10²¹ molecules

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To solve this problem, we have to work from the known to the unknown specie;

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                  6.02 x 10²³ molecules of water  = 1 mole

                   1.2 x 10²¹ molecules of water  = \frac{1.2 x 10^{21} }{6.02 x 10^{23} }    = 0.2 x 10⁻²moles

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           0.002 moles of water will be produced from \frac{0.002}{2}  = 0.001moles

To find the mass;

           Mass  = number of moles x molar mass

Molar mass of  SnO₂ = 118.7 + 2(16) = 150.7g/mol

        Mass  =  0.001 x 150.7 = 0.15g

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