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? 2 K + CrBr2 2 KBr + Cr

Mass of Chromium produced = 37.91 grams

Explanation:

2K + CrBr₂ → 2KBr + Cr

2mole 1 mole 1 mole

mass of Potassium = 57.0 grams

molar mass of Potassium = 39.1 g/mol

no of moles of Potassium = 57.0 / 39.1 = 1.458 moles

mass of CrBr₂ = 199 grams

molar mass of CrBr₂ = 211.8 gram/mole

no of moles of CrBr₂ = 199 / 211.8 = 0.939 mole

From chemical equation

1 mole of CrBr₂ = 2 moles of K

∴ 0.939 moles of CrBr₂ = ?

⇒ 0.939 x 2/1 = 1.878 moles of K

1.878 moles of K is needed, but there is 1.458 moles of K. So, Potassium is completed first during the reaction. Hence, Potassium is limiting reagent. and CrBr₂ is excess reagent.

From chemical equation

2 moles of K = 1 mole of Cr

∴ 1.458 moles of K = ?

⇒ 1.458 x 1 / 2 = 0.729 moles of Cr

no of moles of Cr formed = 0.729 moles

molar mass of Cr = 52.0 g/mol

mass of one mole of Cr = 52.0 grams

mass of 0.729 moles of Cr = 52.0 x 0.729 = 37.908 grams

mass of Chromium produced = 37.91 grams