1 g Ca is burnt with 1 g O2 in a closed vessel. Which reactant is left in excess and how much? (Given; atomic weight: Ca = 40; O = 16)0.4 g Ca0.2 g O20.6 g O20.2 g Ca

First, we need to determine the number of moles for each reactant.

The number of moles is calculated by dividing the mass by the atomic weight.

For Calcium (Ca): 1 g / 40 g/mol = 0.025 mol

For Oxygen (O2): 1 g / 32 g/mol = 0.03125 mol

The balanced chemical equation for the reaction of calcium with oxygen is: 2Ca + O2 -> 2CaO

From the balanced equation, we can see that the mole ratio of Ca to O2 is 2:1. This means that for every 2 moles of Ca, we need 1 mole of O2.

However, we have 0.025 mol of Ca and 0.03125 mol of O2. This means we have more than enough O2 for the reaction (because we only need 0.0125 mol of O2 for 0.025 mol of Ca), and therefore, Ca is the limiting reactant.

To find out how much O2 is left, we subtract the amount of O2 that reacted with Ca from the total amount of O2:

0.03125 mol - 0.0125 mol = 0.01875 mol

To convert this back to grams, we multiply by the molar mass of O2:

0.01875 mol * 32 g/mol = 0.6 g

So, O2 is left in excess, and the amount left is 0.6 g.