Question

Stainless steel is an alloy composed of between 11 and 30% chromium by mass. It may also contain other elements, such as nickel or manganese. The alloy is rust-resistant. A stainless steel alloy is to be analyzed for its chromium content. A 3.50 g sample of the steel is used to produce 250.0 ml of a solution containing Cr2O7²-. A 10.0-ml portion of this solution is added to BaCl₂(aq). When the pH of the solution is properly adjusted, 0.145 g BaCrO₄(s) precipitates. What might be the approximate percentage of chromium in the steel?

A) 12.1%
B) 14.3%
C) 16.7%
D) 18.5%

Answer 1

To determine the approximate percentage of chromium in the steel sample, the mass of barium chromate precipitated is used to calculate the moles of chromium, which are then related to the original sample mass after accounting for solution dilutions. The calculated percentage of chromium does not match the provided answer choices exactly, but 18.5% (D) is the closest option.

Step-by-step explanation:

The question involves determining the approximate percentage of chromium in the steel from a stainless steel alloy sample analysis, where the steel alloy contains chromium as an essential corrosion-resistant element. To find the percentage of chromium in the sample, one must first determine the molar mass of barium chromate (BaCrO4), which is 253.32 g/mol. The mass of BaCrO4 precipitated (0.145 g) is then used to calculate the moles of chromate ions (and thus chromium ions since they are in a 1:1 ratio in BaCrO4), and these moles are mathematically related to the original 3.50 g sample, considering the dilution factor introduced by the preparation of the 250.0 ml solution and the subsequent 10.0 ml aliquot taken for the BaCl2 reaction.

With the molar mass of chromium being 51.996 g/mol and one mole of Cr forming one mole of BaCrO4, we can calculate the moles of Cr in the 0.145 g of BaCrO4 by the following steps:

The moles of BaCrO4 = Mass of BaCrO4 / Molar mass of BaCrO4 = 0.145 g / 253.32 g/mol = 0.000572 mol

Since the moles of BaCrO4 are equal to the moles of Cr, moles of Cr = 0.000572 mol.

The total moles of Cr in the 250.0 ml solution then are 0.000572 mol x (250.0 ml / 10.0 ml) = 0.0143 mol. To find the grams of Cr in the initial 3.50 g sample, we multiply the moles of Cr by its molar mass: 0.0143 mol x 51.996 g/mol = 0.744 g.

Finally, to find the percentage of chromium in the initial steel sample, we divide the mass of Cr by the mass of the steel sample and multiply by 100: (0.744 g Cr / 3.50 g sample) x 100 = approximately 21.3%.

This is not an exact answer to choose from, but it should be closest to 18.5% (D), suggesting there might be a discrepancy or error in the question or answer choices given.