Question

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

Answer 1

0.252 g O₂

Step-by-step 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.

• We would also need 2 zincs on both sides if we balance oxygen

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

`1.28 \hspace{3} g \hspace{3} ZnO(\frac{1 \hspace{3} mol \hspace{3} ZnO}{81.39 \hspace{3} g \hspace{3} ZnO} )(\frac{1 \hspace{3} mol \hspace{3} O_2}{2 \hspace{3} mol \hspace{3} ZnO} )(\frac{32.00 \hspace{3} g \hspace{3} O_2}{1 \hspace{3} mol \hspace{3} O_2} )` = 0.251628 g O₂

Step 5: Simplify

We are given 3 sig figs.

0.251628 g O₂ ≈ 0.252 g O₂