Answer:
The answer to your question is 0.113 moles of Fe₂O₃
Explanation:
Data
moles of Fe₂O₃ = ?
mass of Fe₂O₃ = 18 grams
Process
1.- Calculate the molar mass of Fe₂O₃
Fe₂O₃ = (56 x 2) + (16 x 3)
= 112 + 48
= 160 g
2.- Use proportions to solve this problem. The molar mass is equivalent to 1 mol.
160 g of Fe₂O₃ --------------- 1 mol
18 g of Fe₂O₃ ---------------- x
x = (18 x 1)/160
x = 0.113 moles of Fe₂O₃
Answer:
Molarity is a unit that measures how much moles of solute dissolved in a liter of solvent. Molarity expressed using capital M while molarity, a different unit, expressed using lower case m.
We want to make 0.005 M solution which means we need 0.005 moles of KmnO4 per liter of water. First, we have to calculate how many grams of KMnO4 we need for the solution.
We want to make 250ml solution, so the number of moles of KMnO4 we need will be: 0.005 mol/liter *(250 ml * 1liter/1000ml)= 0.005 mol/liter * 1/4 liter = 0.00125 moles
The molecular mass of KMnO4 is 158g/mol, so the mass of KMnO4 we need will be: 0.00125 moles * 158g/mol= 0.1975 grams
We know that we need 0.1975 g of KMnO4, now we weigh them and put it inside a dish. After that, we prepare Erlenmeyer or a volumetric flask filled with water half of the volume needed(125ml). Pour the weighted solute into the flask, stir until all solute dissolved.
Then we add water to the container slowly until its volume reaches the 250ml mark.
The balance chemical equation is follow,
2 I⁻ + SO₄²⁻ + 4 H⁺ → I₂ + SO₂ + 2 H₂O
According to this reaction, 2 moles of I⁻ reacts with 1 mole of SO₄²⁻ to produce 1 mole of I₂ and 1 mole of SO₂.
Result:
So with the formation of 1 mole of I₂, 1 mole of SO₂ is produced.
Answer:
677.39 g/mol
Explanation:
Step 1: Find molar masses of elements
Carbon (C) - 12.01 g/mol
Hydrogen (H) - 1.01 g/mol
Oxygen (O) - 16.00 g/mol
Step 2: Multiply the amount present
12.01 g/mol · 46 = 552.46 g/mol
1.01 g/mol · 92 = 92.92 g/mol
16.00 g/mol · 2 = 32.00 g/mol
Step 3: Add up all the molar masses to find compound molar mass
552.46 g/mol + 92.92 g/mol + 32.00 g/mol = 677.39 g/mol
