Answer: 1. is the limiting reagent and is the excess reagent.
2. 2.91 g of excess reagent
3. 4.66 g of carbon dioxide produced.
Explanation:
To calculate the moles :
According to stoichiometry :
2 moles of require = 1 mole of
Thus 0.106 moles of will require= of
Thus is the limiting reagent as it limits the formation of product and is the excess reagent.
Amount of excess reagent = ( 0.144-0.053) moles = 0.091 moles
Mass of excess reagent
As 2 moles of give = 2 moles of
Thus 0.106 moles of give = of
Mass of