10HI+2KMNO4+3H2SO4-5I2+2MNSO4+K2SO4+8H20
Find the mass of water produced when 300g HI reacts with 30 g KMno4 and excess H2So4.
Find the mass of k2So4 produces 30 mL of .250 M KMno4 reacts with 400 mL .100 M HI.
mols HI = grams/molar mass = ?
mols KMnO4 = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols HI to mols H2O.
Same procedure convert mols KMnO4 to mols H2O.
It is likely that the values for mol;s H2O will be different which means one of them must be wrong. In limiting reagent problems, which this is, the smaller values is ALWAYS the correct one and the reagent producing that number is the limiting reagent.
Using the smaller value, convert to grams H2O. grams = mols H2O x molar mass H2O
i do not get it why convert grams H2O. grams = mols H2O x molar mass H2O.
Because you already got the smallest value(limiting reagent)
To find the mass of water produced in the reaction between 300g HI and 30g KMnO4 with excess H2SO4, we need to first write and balance the chemical equation:
10HI + 2KMnO4 + 3H2SO4 → 5I2+ 2MnSO4 + K2SO4 + 8H2O
From the balanced equation, we can see that the stoichiometric ratio between HI and H2O is 10:8, which means that for every 10 moles of HI reacted, 8 moles of H2O are produced. To find the mass of water, we can use the following steps:
Step 1: Convert the mass of HI to moles:
Molar mass of HI (hydrogen iodide) = 1.01 + 126.9 = 127.91 g/mol
Moles of HI = Mass of HI / Molar mass of HI
Moles of HI = 300 g / 127.91 g/mol
Step 2: Determine the moles of water produced using stoichiometry:
According to the balanced equation, 10 moles of HI produce 8 moles of H2O.
Moles of H2O = Moles of HI x (8 moles H2O / 10 moles HI)
Step 3: Convert the moles of water to grams:
Molar mass of H2O (water) = 2.02 + 16.00 = 18.02 g/mol
Mass of H2O = Moles of H2O x Molar mass of H2O
Now we can calculate the mass of water produced:
Mass of HI:
Moles of HI = 300 g / 127.91 g/mol ≈ 2.3448 mol
Moles of H2O:
Moles of H2O = 2.3448 mol x (8 mol H2O / 10 mol HI) ≈ 1.875 mol
Mass of H2O:
Mass of H2O = 1.875 mol x 18.02 g/mol ≈ 33.77 g
Therefore, the mass of water produced when 300g HI reacts with 30g KMnO4 and excess H2SO4 is approximately 33.77 grams.
Moving on to the second question:
To find the mass of K2SO4 produced when 30 mL of 0.250 M KMnO4 reacts with 400 mL of 0.100 M HI, we need to first write and balance the chemical equation:
10 HI + 2 KMnO4 + 3 H2SO4 → 5 I2 + 2 MnSO4 + K2SO4 + 8 H2O
From the balanced equation, we can see that the stoichiometric ratio between KMnO4 and K2SO4 is 2:1, which means that for every 2 moles of KMnO4 reacted, 1 mole of K2SO4 is produced. To find the mass of K2SO4, we can use the following steps:
Step 1: Convert the volume of KMnO4 solution to moles:
Molarity (M) = Moles / Volume (L)
Moles of KMnO4 = Molarity x Volume (L)
Moles of KMnO4 = 0.250 M x 0.030 L
Step 2: Determine the moles of K2SO4 produced using stoichiometry:
According to the balanced equation, 2 moles of KMnO4 produce 1 mole of K2SO4.
Moles of K2SO4 = Moles of KMnO4 x (1 mole K2SO4 / 2 moles KMnO4)
Step 3: Convert the moles of K2SO4 to grams:
Molar mass of K2SO4 (potassium sulfate) = 39.10 + (32.07 + 16.00x4) = 174.26 g/mol
Mass of K2SO4 = Moles of K2SO4 x Molar mass of K2SO4
Now we can calculate the mass of K2SO4 produced:
Moles of KMnO4:
Moles of KMnO4 = 0.250 M x 0.030 L ≈ 0.0075 mol
Moles of K2SO4:
Moles of K2SO4 = 0.0075 mol x (1 mol K2SO4 / 2 mol KMnO4) ≈ 0.00375 mol
Mass of K2SO4:
Mass of K2SO4 = 0.00375 mol x 174.26 g/mol ≈ 0.6521 g
Therefore, the mass of K2SO4 produced when 30 mL of 0.250 M KMnO4 reacts with 400 mL of 0.100 M HI is approximately 0.6521 grams.
To find the mass of water produced when 300g HI reacts with 30g KMnO4 and excess H2SO4, you need to balance the chemical equation and determine the stoichiometry of the reaction.
Step 1: Writing the balanced chemical equation:
The balanced chemical equation for the reaction between HI, KMnO4, and H2SO4 is as follows:
10HI + 2KMnO4 + 3H2SO4 => 5I2 + 2MnSO4 + K2SO4 + 8H2O
Step 2: Determining the stoichiometry of the reaction:
From the balanced equation, the stoichiometric coefficient of water (H2O) is 8, which means that for every 8 moles of water produced, we need 10 moles of HI.
Step 3: Convert the given mass of HI to moles:
Using the molar mass of HI (127.9 g/mol), we can calculate the number of moles of HI:
300g HI * (1 mol HI / 127.9 g HI) = 2.34 mol HI
Step 4: Calculate the moles of water produced:
Since the stoichiometric coefficient of water is 8, we can calculate the moles of water produced using the mole ratio:
2.34 mol HI * (8 mol H2O / 10 mol HI) = 1.87 mol H2O
Step 5: Convert moles of water to grams:
Using the molar mass of water (18.015 g/mol), we can convert the moles of water to grams:
1.87 mol H2O * (18.015 g H2O / 1 mol H2O) = 33.67 g H2O
Therefore, when 300g HI reacts with 30g KMnO4 and excess H2SO4, the mass of water produced is 33.67g.
To find the mass of K2SO4 produced when 30 mL of 0.250 M KMnO4 reacts with 400 mL of 0.100 M HI, follow the steps below:
Step 1: Writing the balanced chemical equation:
The balanced chemical equation for the reaction between KMnO4 and HI is the same as before:
10HI + 2KMnO4 + 3H2SO4 => 5I2 + 2MnSO4 + K2SO4 + 8H2O
Step 2: Determining the stoichiometry of the reaction:
From the balanced equation, the stoichiometric coefficient of K2SO4 is 1. This means that for every 2 moles of KMnO4, we can generate 1 mole of K2SO4.
Step 3: Convert the volume of KMnO4 and HI to moles:
Using the molarity (M) and volume (V) values, we can calculate the number of moles of KMnO4 and HI:
KMnO4: 0.250 M * 0.030 L = 0.0075 mol KMnO4
HI: 0.100 M * 0.400 L = 0.040 mol HI
Step 4: Determine the limiting reactant:
To determine the limiting reactant, compare the moles of KMnO4 and HI using their stoichiometric coefficients:
KMnO4: 0.0075 mol KMnO4 * (10 mol HI / 2 mol KMnO4) = 0.0375 mol HI
Since the moles of HI (0.040 mol) are greater than the moles of KMnO4 (0.0375 mol), KMnO4 is the limiting reactant.
Step 5: Calculate the moles of K2SO4 produced:
Using the stoichiometric coefficient of 1 for K2SO4, we can calculate the moles of K2SO4 produced:
0.0075 mol KMnO4 * (1 mol K2SO4 / 2 mol KMnO4) = 0.00375 mol K2SO4
Step 6: Convert moles of K2SO4 to grams:
Using the molar mass of K2SO4 (174.3 g/mol), we can convert the moles of K2SO4 to grams:
0.00375 mol K2SO4 * (174.3 g K2SO4 / 1 mol K2SO4) = 0.654 g K2SO4
Therefore, when 30 mL of 0.250 M KMnO4 reacts with 400 mL of 0.100 M HI, the mass of K2SO4 produced is 0.654g.