Question

Calculate number of g H3PO3 formed from 53.6 g H2O react with excess PCl3?

Answer 1

Answer:

81.3 g H₃PO₃

General Formulas and Concepts:

Math

Pre-Algebra

Order of Operations: BPEMDAS

1. Brackets
2. Parenthesis
3. Exponents
4. Multiplication
5. Division
6. Addition
7. Subtraction

• Left to Right

Chemistry

Atomic Structure

• Reading a Periodic Table

Stoichiometry

• Reaction Molar Ratios
• Using Dimensional Analysis

Explanation:

Step 1: Define

[RxN - Unbalanced] PCl₃ + H₂O → HCl + H₃PO₃

↓

[RxN - Balanced] PCl₃ + 3H₂O → 3HCl + H₃PO₃

[Given] 53.6 g H₂O

[Solve] x g H₃PO₃

Step 2: Identify Conversions

[RxN] 3 mol H₂O → 1 mol H₃PO₃

[PT] Molar Mass of H - 1.01 g/mol

[PT] Molar Mass of O - 16.00 g/mol

[PT] Molar Mass of P - 30.97 g/mol

Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol

Molar Mass of H₃PO₃ - 3(1.01) + 30.97 + 3(16.00) = 82.00 g/mol

Step 3: Stoich

1. [S - DA] Set up:                                                                                               $\displaystyle 53.6 \ g \ H_2O(\frac{1 \ mol \ H_2O}{18.02 \ g \ H_2O})(\frac{1 \ mol \ H_3PO_3}{3 \ mol \ H_2O})(\frac{82.00 \ g \ H_3PO_3}{1 \ mol \ H_3PO_3})$
2. [S - DA] Multiply/Divide [Cancel out unit]:                                                     $\displaystyle 81.3023 \ g \ H_3PO_3$

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

81.3023 g H₃PO₃ ≈ 81.3 g H₃PO₃