"n(CO_2)=\\frac{15.8g}{44\\frac{g}{mol}}=0.359 mol"
When 0.359 mol of "CO_2" react with sufficient amount of "H_2" 8.35 kJ of energy are absorbed.
When 2 moles of "CO_2" react with sufficient amount of "H_2" xkJ of enerhy is absorbed.
Solve the proportion:
"x= 46.52 kJ"
Comments
Leave a comment