a chemist reacted 64 grams of Ru with O2. This reaction produced 160 grams of ruthenium oxide and 4 grams of leftover O2. How much of the reactants would this scientist need to make 320 grams of the product with no leftover O2?
Ru + O2 = RuO2 + O2 leftover
To determine the amount of reactants needed to produce a specific amount of product, we can set up a proportion based on the given information.
We can start by analyzing the given reaction equation:
Ru + O2 = RuO2 + O2 leftover
From the equation, we can see that the stoichiometric ratio of Ru to RuO2 is 1:1. This means that 1 mole of Ru reacts with 1 mole of RuO2. Similarly, the stoichiometric ratio of O2 to RuO2 is 1:1, indicating that 1 mole of O2 reacts with 1 mole of RuO2.
To solve the problem, let's break it down into steps:
Step 1: Convert the given mass of ruthenium oxide (RuO2) to moles.
Given mass of RuO2 = 160 grams
Molar mass of RuO2 = 101.07 g/mol (from the periodic table)
Number of moles of RuO2 = given mass / molar mass = 160 g / 101.07 g/mol
Step 2: Determine the moles of Ru required to react with the moles of RuO2.
Since the stoichiometric ratio for Ru to RuO2 is 1:1, the moles of Ru required will be equal to the moles of RuO2.
Number of moles of Ru = Number of moles of RuO2
Step 3: Calculate the mass of Ru required.
Mass of Ru = number of moles × molar mass = number of moles × molar mass of Ru
Step 4: Set up a proportion to find the mass of Ru required to produce 320 grams of RuO2.
Let x be the mass of Ru required to produce 320 grams of RuO2.
Mass of Ru / Mass of RuO2 = x / 320 g
Using the calculated mass of Ru from Step 3: (mass of Ru) / (160 g) = x / (320 g)
Now we can solve for x by cross-multiplying and solving the equation for x.
Solving the equation will give us the amount of Ru required to produce 320 grams of RuO2 with no leftover O2.