Question

Помогите пожалуйста. Даю 30 балов На одну чашку весов положили медные стружки количеством вещества 5,6 моль. Какая масса должна быть у порции оксида железа(III) чтобы весы были уравновешены? Помогите пожалуйста. Даю 20 балов

Answer 1

M(Cu)=64*5.6=358.4 г
значит и Fe2O3 m=358.4 г

Answer 2

To determine the mass of the portion of iron(III) oxide needed to balance the scales, we need to use the concept of molar ratios and the molar mass of iron(III) oxide.

Molar Ratios:

The balanced chemical equation for the reaction between copper and iron(III) oxide is:

2Fe2O3 + 3Cu -> 3CuO + 4Fe

From the balanced equation, we can see that 2 moles of iron(III) oxide react with 3 moles of copper. Therefore, the molar ratio between iron(III) oxide and copper is 2:3.

Molar Mass:

The molar mass of iron(III) oxide (Fe2O3) can be calculated by adding up the atomic masses of its constituent elements. The atomic masses of iron (Fe) and oxygen (O) are approximately 55.85 g/mol and 16.00 g/mol, respectively.

Molar mass of Fe2O3 = (2 * atomic mass of Fe) + (3 * atomic mass of O) Molar mass of Fe2O3 = (2 * 55.85 g/mol) + (3 * 16.00 g/mol) Molar mass of Fe2O3 = 159.70 g/mol

Calculation:

To find the mass of iron(III) oxide needed to balance the scales, we can set up a proportion using the molar ratio:

(5.6 mol Cu) / (3 mol Fe2O3) = (x g Fe2O3) / (159.70 g Fe2O3)

Solving for x, we can find the mass of iron(III) oxide needed:

x = (5.6 mol Cu * 159.70 g Fe2O3) / (3 mol Fe2O3) x ≈ 299.47 g Fe2O3

Therefore, the mass of the portion of iron(III) oxide needed to balance the scales is approximately 299.47 grams.

Please note that the given information does not specify the molar mass of copper, so we cannot calculate the exact mass of iron(III) oxide without that information.