Answer:
1822.72 g
Explanation:
Applying,
n = R.M/M.M.................. Equation 1
Where n = number of moles of iron(II) nitrate, R.M = Reacting mass of iron(II) nitrate, M.M = molar mass of Iron(II) nitrate.
Make R.M the subject of the equation
R.M = n×M.M............. Equation 2
From the question,
Given: n = 6.33 mol
Constant: M.M of iron(II) nitrate = 287.95 g/mol
Substitute these values into equation 2
R.M = 6.33(287.95)
R.M = 1822.72 g
Hence the mass of iron(II) nitrate is 1822.72 g
Answer:
We would need 10 mL of the concentrated CaCl₂ stock solution, and 30 mL of water.
Explanation:
To solve the question asked we can use the C₁V₁=C₂V₂ equation, where:
We <u>solve for V₁</u>:
We would need 10 mL of the concentrated CaCl₂ stock solution, and (40-10) 30 mL of water.
Answer:
1.21 mol KClO₃
General Formulas and Concepts:
<u>Atomic Structure</u>
- Reading a Periodic Table
- Moles
- Mole Ratio
<u>Stoichiometry</u>
- Analyzing reactions rxn
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
<em>Identify variables</em>
[rxn] 2KClO₃ → 2KCl + 3O₂
[Given] 58.3 g O₂
[Solve] mol KClO₃
<u>Step 2: Identify Conversions</u>
[rxn] 2 mol KClO₃ → 3 mol O₂
[PT] Molar Mass of O: 16.00 g/mol
Molar Mass of O₂: 2(16.00) = 32.00 g/mol
<u>Step 3: Convert</u>
- [DA] Set up:

- [DA] Divide/Multiply [Cancel out units]:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
1.21458 mol KClO₃ ≈ 1.21 mol KClO₃