Answer:
Mass of Chromium produced = 37.91 grams
Step-by-step 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