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What mass of chromium would be produced from the reaction of 57.0 g of potassium with 199 g of chromium(II) bromide according to the following reaction? Which reactant is in excess, and how many grams of this reactant are left over?

2 K + CrBr₂ → 2 KBr + Cr

a) 84.0 g, potassium is in excess, 57.0 g left over
b) 89.0 g, chromium(II) bromide is in excess, 23.0 g left over
c) 84.0 g, chromium(II) bromide is in excess, 23.0 g left over
d) 89.0 g, potassium is in excess, 57.0 g left over

User Nene
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Final answer:

The mass of chromium produced from the given amounts of potassium and chromium(II) bromide, using stoichiometry, is 37.882 g. None of the answer choices provided match this calculated value. Potassium is the limiting reactant, leaving an excess of 41.7513 g of chromium (II) bromide.

Step-by-step explanation:

To determine the mass of chromium produced from the reaction of potassium with chromium (II) bromide, we need to use stoichiometry based on the balanced chemical equation:

2 K + CrBr₂ → 2 KBr + Cr

Firstly, we calculate the moles of the reactants. The molar mass of potassium (K) is approximately 39.1 g/mol, and for chromium (II) bromide (CrBr₂), it's approximately 215.8 g/mol. By dividing the given masses by these molar masses, we get the moles of each reactant:

  • Potassium: 57.0 g / 39.1 g/mol = 1.457 moles of K
  • Chromium (II) bromide: 199 g / 215.8 g/mol = 0.922 moles of CrBr₂

According to the balanced equation, the reaction requires 2 moles of K for every mole of CrBr₂. Therefore, we divide the moles of potassium by 2 to find out how much Cr it can react with:

1.457 moles of K / 2 = 0.7285 moles of CrBr₂

Since 0.7285 moles of K can react with 0.7285 moles of CrBr₂, and we only have 0.922 moles of CrBr₂, we see that potassium is the limiting reactant. We need to find the maximum amount of Cr that can be produced, which will be the same as the moles of chromium (II) bromide that can react with the available potassium:

  • Mass of chromium: 0.7285 moles × 52.0 g/mol (molar mass of Cr) = 37.882 g of Cr

None of the answer choices match the calculated mass. Thus, it seems there's an error with the choices provided. Regarding the excess reactant, since potassium is the limiting reactant, chromium (II) bromide is in excess. To find out how much is left over:

  • 0.922 moles initial CrBr₂ - 0.7285 moles reacted CrBr₂ = 0.1935 moles CrBr₂ unreacted
  • Mass left over: 0.1935 moles × 215.8 g/mol = 41.7513 g of CrBr₂

As this value does not match any of the answers provided, it seems there might be a mistake in the question or the answer options.

User EsmaeelE
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