1) List the reactants: sodium bicarbonate (NaHCO₃) and citric acid (H₃C₆H₅O₇).
Reactants undergo change during a chemical reaction.
2) List the products: water (H₂O), carbon dioxide (CO₂) and sodium citrate (Na₃C₆H₅O₇).
Products are the substances formed from chemical reactions.
3) The balanced chemical equation:
3NaHCO₃ + H₃C₆H₅O₇ → 3H₂O + 3CO₂ + Na₃C₆H₅O₇.
Answer:
Partial pressure of
in the gas was 733 torr and mass of
in the sample was 2.12 g.
Explanation:
a) Total pressure of gas = (partial pressure of water vapour)+(partial pressure of
)
Here partial pressure of water vapour is 21 torr and total pressure of gas is 754 torr.
So, partial pressure of
= (total pressure of gas)-(partial pressure of water vapour) = (754 torr) - (21 torr) = 733 torr
b) Lets assume that
behaves ideally. Hence-
PV=nRT
where P is pressure of
, V is volume of
, n is number of moles of
, R is gas constant and T is temperature in kelvin
here P = 733 torr =
= 0.9646 atm
V = 0.65 L, R = 0.082 L.atm/(mol.K), T=(273+22)K = 295 K
So, 
= 
= 0.0259 moles
As 3 moles of
are produced from 2 moles of
therefore 0.0259 moles of
are produced from
moles or 0.0173 moles of
.
Molar mass of
= 122.55 g
So mass of
in sample = 
= 2.12 g
<h3>
Answer:</h3>
7.4797 g AlF₃
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Stoichiometry</u>
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN] 2AlF₃ + 3K₂O → 6KF + Al₂O₃
[Given] 15.524 g KF
<u>Step 2: Identify Conversions</u>
[RxN] 2 mol AlF₃ = 6 mol KF
Molar Mass of Al - 26.98 g/mol
Molar Mass of F - 19.00 g/mol
Molar Mass of K - 39.10 g/mol
Molar Mass of AlF₃ - 26.98 + 3(19.00) = 83.98 g/mol
Molar Mass of KF - 39.10 + 19.00 = 58.10 g/mol
<u>Step 3: Stoichiometry</u>
- Set up:

- Multiply/Divide:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 5 sig figs.</em>
7.47966 g AlF₃ ≈ 7.4797 g AlF₃