How many grams of oxygen gas must react to give 1.28 g of ZnO:

0.252 g O₂

Explanation:

Step 1: Find RxN

O₂ (g) + Element X (s) →  ZnO (s)

O₂ (g) + Zn (s) → ZnO (s)

Step 2: Balance RxN

We need 2 oxygens on both sides.

O₂ (g) + 2Zn (s) → 2ZnO (s)

Step 3: Define

Given - 1.28 g ZnO

Find - x g O₂

Molar Mass of Zn - 65.39 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of O₂ - 2(16.00) = 32.00 g/mol

Molar Mass of ZnO - 65.39 + 16.00 = 81.39 g/mol

Step 4: Stoichiometry

= 0.251628 g O₂

Step 5: Simplify

We are given 3 sig figs.

0.251628 g O₂ ≈ 0.252 g O₂

The structure of ammonium lauryl sulfate is described below in complete details.

Explanation:

The compound comprises ammonium ions and lauryl sulfate. Lauryl sulfate includes lauric acid (in white and black), the fatty acid produced by the covalent bonds among C-C connected to hydrogen, and lauric acid (in red) between C-S covalent bond connected to sulfate ions. oxygen is connected to Sulfur through covalent bonds. In Ammonium ions, N is enclosed by four hydrogen atoms.