# 130 g of a mixture of the three platinum metals A, B and C were crushed and dissolved in aqua regia.

**130 g of a mixture of the three platinum metals A, B and C were crushed and dissolved in aqua regia. The total volume of released gas is 26.88 liters. At the same time, metal C, weighing 7.9 g, remained in the sediment. It is known that metal A – reacts with individual oxidizing acids and one volume of metal A absorbs up to 700 volumes of hydrogen. Metal B is the main component of the alloy from which the kilogram standard is made. Metal C is the strongest among the simple substances. Accept that when aqua regia interacts with metals, elements A and B form complex compounds in which they are in the same oxidation state. Establish what metals are in question. In the first field, enter the characters of the elements A-C sequentially, separated by commas, without spaces. In the following fields, indicate the mass fraction of metals in the mixture (the order does not matter). Indicate your answer to the nearest tenths, in percent, without specifying the dimension.**

Given:

m mixture (A, B, C) = 130 g

V (gas) = 26.88 l

m (C) = 7.9 g

To find:

ω (A) -?

ω (B) -?

ω (C) -?

Solution:

1) Metal A – palladium (Pd), metal B – platinum (Pt), metal C – iridium (Ir);

2) 3Pd + 18HCl + 4HNO3 => 3H2 [PdCl6] + 4NO ↑ + 8H2O;

3Pt + 18HCl + 4HNO3 => 3H2 [PtCl6] + 4NO ↑ + 8H2O;

3) m mixture (A, B) = m mixture (A, B, C) – m (C) = 130 – 7.9 = 122.1 g;

4) Let m (Pd) = (x) r, then m (Pt) = (122,1 – x) r;

5) n (Pd) = m (Pd) / M (Pd) = (x / 106) mol;

6) n1 (NO) = n (Pd) * 4/3 = (x / 106) * 4/3 = (4x / 318) mol;

7) n (Pt) = m (Pt) / M (Pt) = ((122.1 – x) / 195) mol;

8) n2 (NO) = n (Pt) * 4/3 = ((122.1 – x) / 195) * 4/3 = ((488.4 – 4x) / 585) mol;

9) n total (NO) = V (NO) / Vm = 26.88 / 22.4 = 1.2 mol;

10) n total (NO) = n1 (NO) + n2 (NO);

1.2 = (4x / 318) + ((488.4 – 4x) / 585);

223236 = 2340x + 155311.2 – 1272x;

1068x = 67924.8;

x = 63.6;

m (Pd) = (x) = 63.6 g;

11) ω (Pd) = m (Pd) * 100% / m mixture (A, B, C) = 63.6 * 100% / 130 = 48.9%;

ω (Ir) = m (Ir) * 100% / m mixture (A, B, C) = 7.9 * 100% / 130 = 6.1%;

ω (Pt) = 100% – ω (Pd) – ω (Ir) = 100% – 48.9% – 6.1% = 45%.

Answer: The mass fraction of Pd is 48.9%; Pt – 45%; Ir – 6.1%.