Question

When rubidium metal is exposed to air, two atoms of rubidium, Rb, combine with one atom
of oxygen. If 1.98 grams of rubidium is exposed to air, what will be the mass of the product
in grams

Answer 1

2.1653 g

Explanation:

The molar mass of Rubidium is;

85.468 g/mol

Therefore the moles of Rubidium that reacted with oxygen is;

1.98 / 85.468

= 0.0232 moles

If every two moles of Rubidium reacts with one mole of oxygen then the amount of oxygen consumed in the chemical reaction is;

0.5 * 0.0232

= 0.0116 moles

The molar mass of an oxygen atom is 16 g/mole. Then the amount of O in grams consumed is;

0.0116 * 16

=0.1853 g

The final weight of the Rubidium II Oxide is;

1.98 + 0.1853

= 2.1653 g

Answer 2

Taking into account the reaction stoichiometry, 2.165 grams Rb₂O are formed when 1.98 grams of rubidium is exposed to air.

Reaction stoichiometry

In first place, the balanced reaction is:

2 Rb + O  → Rb₂O

By reaction stoichiometry (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:

• Rb: 2 moles
• O: 1 mole  
• Rb₂O:  1 mole

The molar mass of the compounds is:

• Rb: 85.47 g/mole
• O₂: 16 g/mole
• Rb₂O: 186.94 g/mole

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

• Rb: 2 moles ×85.47 g/mole= 170.94 grams
• O₂: 1 mole ×16 g/mole= 16 grams
• Rb₂O: 1 mole ×186.94 g/mole= 186.94 grams

Mass of Rb₂O formed

The following rule of three can be applied: if by reaction stoichiometry 170.94 grams of Rb form 186.94 grams of Rb₂O, 1.98 grams of Rb form how much mass of Rb₂O?

$mass of Rb_{2} O=\frac{1.98 grams of Rbx186.94 grams of Rb_{2}O }{170.94 grams of Rb}$

mass of Rb₂O= 2.165 grams

Then, 2.165 grams Rb₂O are formed when 1.98 grams of rubidium is exposed to air.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699