Im really struggling !

2au2o3 > 4au + 3o2

only place the numerical value in the answer spot.

round your final answer to the nearest whole number.

use the equation above to answer the following questions:

1. what volume of o2 at stp is produced from the reaction of 212 grams of au2o3?

2. how many grams of au are produced?

3. how much au2o3 is needed to form 85 grams gold?

1- 16.0 L.

2- 189.0 g.

3- 95.0 g.

Explanation:

1. What volume of O₂ at STP is produced from the reaction of 212 grams of Au₂O₃?

To solve this problem, we use the balanced equation:

2Au₂O₃ → 4Au + 3O₂

It is clear that 2.0 moles of Au₂O₃ are decomposed to give 4.0 moles of Au and 3.0 moles of O₂.Firstly, we should convert the grams of Au₂O₃ (212.0 g) reacted into no. of moles using the relation: n = mass / molar mass.The no. of moles of Au₂O₃ = mass / molar mass = (212.0 g) / (441.93 g/mol) = 0.479 mol.

Using cross multiplication:

2.0 moles of Au₂O₃ produces → 3.0 moles of O₂, from the balanced equation.

0.479 mole of Au₂O₃ produces → ??? moles of O₂.

∴ The no. of moles of O₂ produced from the decomposition of 212.0 g of Au₂O₃ = (0.479 mol)(3.0 mol) / (2.0 mol) = 0.72 mol.Finally, we can get the volume of O₂ produced using the gas law: PV = nRT.

where, P is the pressure of the gas (P = 1.0 atm at STP),

V is the volume of the gas in L (V = ??? L),

n is the no. of moles of the gas (n = 0.72 mol),

R id the general gas constant (R = 0.082 L.atm/mol.K),

T is the temperature of the gas (T = 273.0 K at STP).

∴ The volume of O₂ produced = nRT/P = (0.72 mol)(0.082 L.atm/mol.K)(273.0 K) / (1.0 atm) = 16.1 L ≅ 16.0 L.

2. How many grams of Au are produced?

By the same way used in Q1,

Using cross multiplication:

2.0 moles of Au₂O₃ produces → 4.0 moles of Au, from the balanced equation.

0.479 mole of Au₂O₃ produces → ??? moles of Au.

∴ The no. of moles of Au produced from the decomposition of 212.0 g of Au₂O₃ = (0.479 mol)(4.0 mol) / (2.0 mol) = 0.958 mol.Finally, we can get the no. of grams of Au produced using the law: mass = n x atomic mass.The no. of grams of Au produced = n x atomic mass = (0.958 mol)(196.96 g/mol) = 188.69 g ≅ 189.0 g.

3. How much Au₂O₃  is needed to form 85 grams gold?

By the same way and using the stichiometry of the balanced equation:Firstly, we need to get the no. of moles of Au produced (85.0 g) using the law: n = mass / atomic mass.The no. of moles of Au produced n = mass / atomic mass = (85.0 g) / (196.96 g/mol) = 0.43 mol.

Using cross multiplication:

2.0 moles of Au₂O₃ produces → 4.0 moles of Au, from the balanced equation.

??? mole of Au₂O₃ produces → 0.43 moles of Au.

∴ The no. of moles of Au₂O₃ needed to from 85.0 g of Au = (0.43 mol)(2.0 mol) / (4.0 mol) = 0.215 mol.

Finally, we can get the no. of grams of Au₂O₃ needed using the law: mass = n x atomic mass.The no. of grams of Au₂O₃ needed = n x atomic mass = (0.215 mol)(441.93 g/mol) = 95.35 g ≅ 95.0 g.