How would I solve this?
If 1550 kJ of heat are produced by the reaction below, how many grams of MnCl2 (MW = 126 g/mol) are consumed?
MnCl2 + 2HNO3 → Mn(NO3)2 + 2HCl + 300 kJ
- AshLv 71 month ago
moles of MnCl2 consumed = (moles of MnCl2 from balanced equation / estimated heat) x actual heat produced
moles of MnCl2 consumed = (1 mol / 300 kJ) x 1550 kJ = 5.17 mol
mass of MnCl2 consumed = 5.17 mol x 126 g/mol = 651 g