How many grams of oxygen gas must react to give...?

A) How many grams of oxygen gas must react to give 1.60g of ZnO?
2Zn(s) O2(g)------>2ZnO(s)
Express ans. with appropriate units.
molar mass ZnO = 81.41
so eqtn tells you
2 moles = 2 x 81.41g ZnO comes from 1 mole = 32g O2
so 1.6g ZnO comes from [32/162.81] x 1.6g O2
use calculator to get answer

B) Assuming all gases are at the same temperature and pressure, how many milliliters of oxgen gas must react to give 1.05 L of Cl2O3?
Cl2(g) O2(g)----->Cl2O3(g)
Use Avogadro's Law that moles = volumes for gases at same temp and pressure
Eqtn is NOT balanced
2Cl2(g) 3O2(g)----->2Cl2O3(g)

2mole = 2vol Cl2O3 comes from 1 mole= 13vol O2
so you need 1.05L x 3/2 = 1.575L = 1575mL O2

(A)
2Zn(s) + O₂(g) → 2ZnO(s)

Method 1 :
Molar mass of ZnO = 65.38 + 16.00 = 81.38 g/mol
n(ZnO) = (1.60 g) / (81.38 g/mol) = 0.01966 mol

Refer to the equation. 1 mole of O₂ reacts to give 2 moles of ZnO.
n(O₂) = 0.01966 × (1/2) = 0.00983 mol

Molar mass of O₂ = 16.00 × 2 = 32.00 g
m(O₂) = 0.00983 × 32.00 = 0.315 g


Method 2 :
Mass fraction of O in ZnO = 16/(65.38 + 16.00) = 16.00/81.38
m(O₂) = m(O in ZnO) = 1.60 × (16.00/81.38) = 0.315 g


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(B)
At the same temperature and pressure, mole ratio of gases is equal to volume ratio.

2Cl₂(g) + 3O₂(g) → 2Cl₂O₃(g)

Refer to the equation.
3 volumes of O₂ reacts with 2 volumes of Cl₂O₃ at the same temperature and pressure.
V(O₂) = (1.05 L) × (3/2) = 1.575 L


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(C)
2NO(g) + O₂(g) → 2NO₂(g)

Refer to the equation. 2 moles of NO reacts with 1 mole of O₂.

When 2 moles of NO completely reacts :
n(O₂) needed = (2 mol) × (1/2) = 1 mol < 1.4 mol
Hence, O₂ is in excess, and NO is the limiting reactant.

...... the answer is : c) NO


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(D)
NO is the limiting reactant.
n(NO) reacts = 2.0 mol

Refer to the equation. 1 mole of NO produces 1 mole of NO₂.
n(NO₂) produced = 2.0 mol