Mass relationships in chemical reaction. Make a solution
Consider the reaction
KOH + CO2 -----> KHCO3
How many grams of KOHCO3 are produced from the complete reaction of 75.9 KOH?
The balanced equation tells us that the molar ratio of KOH to KHCO3 is 1:1. This means that for every 1 mole of KOH consumed, 1 mole of KHCO3 is produced.
To determine the amount of KHCO3 produced from 75.9 g of KOH, we need to first convert grams to moles using the molar mass of KOH:
75.9 g KOH x (1 mol KOH / 56.11 g KOH) = 1.35 mol KOH
Since the molar ratio of KOH to KHCO3 is 1:1, we know that 1.35 mol of KHCO3 will be produced.
To convert moles of KHCO3 to grams, we use the molar mass of KHCO3:
1.35 mol KHCO3 x (100.11 g KHCO3 / 1 mol KHCO3) = 135 g KHCO3
Therefore, 75.9 g of KOH will produce 135 g of KHCO3 in this reaction.
To find out the number of grams of KHCO3 produced from the complete reaction of 75.9 grams of KOH, we need to use the concept of stoichiometry and mass relationships in chemical reactions.
Step 1: Write out the balanced chemical equation:
KOH + CO2 ---> KHCO3
Step 2: Determine the molar mass of KOH:
KOH = (39.10 g/mol K) + (16.00 g/mol O) + (1.01 g/mol H)
= 56.11 g/mol
Step 3: Calculate the number of moles of KOH using the given mass and molar mass:
75.9 g KOH * (1 mol KOH / 56.11 g KOH) = 1.354 mol KOH
Step 4: Examine the stoichiometry of the balanced equation. From the equation, we can see that the ratio of KOH to KHCO3 is 1:1.
Step 5: Using the stoichiometry, we can determine the number of moles of KHCO3 produced:
1.354 mol KOH * (1 mol KHCO3 / 1 mol KOH) = 1.354 mol KHCO3
Step 6: Finally, calculate the mass of KHCO3 produced using the molar mass of KHCO3:
KHCO3 = (39.10 g/mol K) + (12.01 g/mol C) + (16.00 g/mol O) + (3 * 16.00 g/mol H)
= 100.11 g/mol
Mass of KHCO3 = 1.354 mol KHCO3 * (100.11 g/mol KHCO3) ≈ 135.8 g
Therefore, approximately 135.8 grams of KHCO3 are produced from the complete reaction of 75.9 grams of KOH.
To determine the number of grams of KOHCO3 produced from the complete reaction of 75.9 grams of KOH, we need to use stoichiometry. Stoichiometry is a method used to calculate the quantities of reactants and products in a chemical reaction.
In this case, we have the balanced chemical equation:
KOH + CO2 -----> KHCO3
First, let's determine the molar mass of each substance involved in the reaction:
- KOH: 39.1 g/mol (atomic mass of potassium (K) = 39.1 g/mol + atomic mass of oxygen (O) = 16 g/mol + atomic mass of hydrogen (H) = 1 g/mol)
- CO2: 44.01 g/mol (atomic mass of carbon (C) = 12.01 g/mol + 2 atomic masses of oxygen (O) = 2 × 16 g/mol)
Next, we need to identify the stoichiometric ratio between KOH and KOHCO3. From the balanced equation, we can see that the ratio of KOH to KOHCO3 is 1:1, meaning that 1 mole of KOH reacts with 1 mole of KOHCO3.
Now, let's calculate the number of moles of KOH:
Given mass of KOH = 75.9 g
Molar mass of KOH = 39.1 g/mol
Number of moles of KOH = Given mass / Molar mass
Number of moles of KOH = 75.9 g / 39.1 g/mol
After calculating this, we find that the number of moles of KOH is approximately 1.94 mol.
Since the ratio of KOH to KOHCO3 is 1:1, we can conclude that the number of moles of KOHCO3 formed is also 1.94 mol.
Finally, let's calculate the mass of KOHCO3:
Molar mass of KOHCO3 = Molar mass of KHCO3 = (atomic mass of K = 39.1 g/mol + atomic mass of H = 1 g/mol + atomic mass of C = 12.01 g/mol + 3 atomic masses of O = 3 × 16 g/mol)
The molar mass of KOHCO3 is approximately 100.1 g/mol.
Mass of KOHCO3 = Number of moles × Molar mass
Mass of KOHCO3 = 1.94 mol × 100.1 g/mol
After calculating this, we find that the mass of KOHCO3 produced from the complete reaction of 75.9 g of KOH is approximately 194 grams.