consider the reaction in which 410g of Ca(NO3)2 react with just the right amount of lithium metal in a single replacement reaction ?
Find out:
a) How many grams of lithium are required?
b) How many grams of each product can be produced?
Here is a solved example of a stoichiometry problem. Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To determine the steps for the given reaction, we first need to write the balanced equation for the reaction between calcium nitrate (Ca(NO3)2) and lithium (Li) metal.
The balanced equation for the reaction is:
3Li + Ca(NO3)2 → 3LiNO3 + Ca
Step 1: Convert the given mass of Ca(NO3)2 to moles.
The molar mass of Ca(NO3)2 is:
Ca: 40.08 g/mol
N: 14.01 g/mol (there are 2 nitrogen atoms)
O: 16.00 g/mol (there are 6 oxygen atoms)
Molar mass of Ca(NO3)2:
= (1 * 40.08) + (2 * 14.01) + (6 * 16.00)
= 40.08 + 28.02 + 96.00
= 164.10 g/mol
Number of moles of Ca(NO3)2:
= Mass / Molar mass
= 410 g / 164.10 g/mol
≈ 2.50 mol
Step 2: Determine the limiting reactant.
To find the limiting reactant, we compare the mole ratio of Ca(NO3)2 to Li in the balanced equation.
From the balanced equation, we see that 3 moles of Li react with 1 mole of Ca(NO3)2.
Number of moles of Li reacted = 3 * number of moles of Ca(NO3)2
= 3 * 2.50 mol
= 7.50 mol
Since we have an excess of Li (as it is given that there is "just the right amount"), the limiting reactant is Ca(NO3)2.
Step 3: Calculate the mass of product formed.
From the balanced equation, we see that the mole ratio of Ca(NO3)2 to Ca is 1:1.
So, the number of moles of Ca formed is also 2.50 mol.
The molar mass of Ca is 40.08 g/mol.
Mass of Ca formed = Number of moles * Molar mass
= 2.50 mol * 40.08 g/mol
= 100.20 g
Therefore, in the reaction, 410 g of Ca(NO3)2 would react to form 100.20 g of Ca.
To determine the products formed and the amount of products in a chemical reaction, we need to balance the equation and use stoichiometry.
The balanced chemical equation for the reaction between calcium nitrate (Ca(NO3)2) and lithium metal (Li) can be written as follows:
2Li + Ca(NO3)2 → 2LiNO3 + Ca
First, we need to convert the given mass of Ca(NO3)2 into moles. To do this, we'll use the molar mass of Ca(NO3)2:
Molar mass of Ca(NO3)2 = (1 mol Ca) + (2 mol N) + (6 mol O)
= 40.08 g/mol + (2 × 14.01 g/mol) + (6 × 16.00 g/mol)
= 164.08 g/mol
Now, we can use the molar mass to calculate the number of moles of Ca(NO3)2:
Moles of Ca(NO3)2 = Mass of Ca(NO3)2 / Molar mass of Ca(NO3)2
= 410 g / 164.08 g/mol
≈ 2.499 mol (rounded to three significant figures)
From the balanced equation, we can see that 2 moles of Li react with 1 mole of Ca(NO3)2. Therefore, the number of moles of Li required is half the number of moles of Ca(NO3)2:
Moles of Li = 0.5 × Moles of Ca(NO3)2
= 0.5 × 2.499 mol
= 1.250 mol
Since the reaction occurs with the balanced stoichiometry, we can conclude that 1.250 moles of Li will be consumed, and an equal amount of LiNO3 will be produced.
To calculate the mass of LiNO3 produced, we'll use its molar mass:
Molar mass of LiNO3 = (1 mol Li) + (1 mol N) + (3 mol O)
= 6.94 g/mol + 14.01 g/mol + (3 × 16.00 g/mol)
= 68.94 g/mol
Mass of LiNO3 = Moles of LiNO3 × Molar mass of LiNO3
= 1.250 mol × 68.94 g/mol
≈ 86.17 g (rounded to two decimal places)
Therefore, approximately 86.17 grams of LiNO3 will be produced in the reaction.