Question

Stoichiometric Calculations: In the following balanced reaction, metallic rubidium reacts with chlorine gas to produce rubidium chloride. 2 Rb (s) + Cl₂ (8) → 2 RbCl (s) A) If 69.27 g of rubidium reacts with 171.2g of chlorine gas, how much rubidium chloride can be produced? (Use proper sig figs)

( )g RbCI

Answer 1

To solve stoichiometric calculations, convert the given masses of reactants into moles according to the balanced equation. In this case, 97.98 g of RbCl can be produced.

Step-by-step explanation:

To solve stoichiometric calculations like this, we need to use the balanced chemical equation and convert the given masses of reactants into moles. From the balanced equation, we see that 2 moles of Rb react with 1 mole of Cl₂ to produce 2 moles of RbCl.

First, let's convert the given mass of Rb into moles. The molar mass of Rb is 85.47 g/mol, so:

69.27 g Rb * (1 mol Rb / 85.47 g Rb) = 0.8101 mol Rb

Next, let's convert the given mass of Cl₂ into moles. The molar mass of Cl₂ is 70.90 g/mol, so:

171.2 g Cl₂ * (1 mol Cl₂ / 70.90 g Cl₂) = 2.415 mol Cl₂

Now, we can compare the mole ratios from the balanced equation. Since the ratio of Rb to Cl₂ is 2:1, and we have 0.8101 mol Rb and 2.415 mol Cl₂, we can only use the smaller value, which is 0.8101 mol Rb. From the ratio, 2 moles of RbCl are produced for every 2 moles of Rb. Therefore, the amount of RbCl produced is:

0.8101 mol Rb * (2 mol RbCl / 2 mol Rb) = 0.8101 mol RbCl

Finally, let's convert the moles of RbCl into grams. The molar mass of RbCl is 120.92 g/mol, so:

0.8101 mol RbCl * (120.92 g RbCl / 1 mol RbCl) = 97.98 g RbCl

Therefore, 97.98 g of RbCl can be produced.