4.67g of calcium reacts with 43.7g of oxygen gas. What is the limiting reactant and the excess reactant. How much of the excess reactant will be left over in grams?
I showed you how to do this in the other problem. This is not a homework site that does all of your homework for you or even practice problems. Just follow how I did the other one to do this one. Post your work if you want me to check it.
The reaction is this.
2Ca + O2 ==> 2CaO
To determine the limiting reactant and the excess reactant, we need to compare the reactants' stoichiometry with the given amounts.
First, we need to write the balanced chemical equation for the reaction between calcium and oxygen:
2Ca + O2 -> 2CaO
From the equation, we can see that the stoichiometric ratio of calcium to oxygen is 2:1. So, for every 2 moles of calcium, we need 1 mole of oxygen.
Next, let's calculate the number of moles of calcium and oxygen using their respective masses and molar masses:
Molar mass of calcium (Ca) = 40.08 g/mol
Molar mass of oxygen (O2) = 32.00 g/mol
Number of moles of calcium = mass of calcium / molar mass of calcium
= 4.67 g / 40.08 g/mol
= 0.1163 mol
Number of moles of oxygen = mass of oxygen / molar mass of oxygen
= 43.7 g / 32.00 g/mol
= 1.3656 mol
Now, we can determine the limiting reactant. Since the stoichiometric ratio of calcium to oxygen is 2:1, we have 0.1163 mol of calcium and 1.3656 mol of oxygen. The ratio is approximately 1:12. Thus, we have an excess amount of oxygen.
To find out the excess amount of oxygen leftover, we can calculate the number of moles of oxygen that would be required to react completely with the given amount of calcium, based on the stoichiometric ratio:
Number of moles of oxygen required = 2 * number of moles of calcium
= 2 * 0.1163 mol
= 0.2326 mol
Excess moles of oxygen = number of moles of oxygen - number of moles of oxygen required
= 1.3656 mol - 0.2326 mol
= 1.133 mol
To find the mass of the excess reactant left over, we can use the molar mass of oxygen:
Mass of excess oxygen = excess moles of oxygen * molar mass of oxygen
= 1.133 mol * 32.00 g/mol
= 36.26 g
Therefore, the limiting reactant is calcium, and the excess reactant is oxygen. The amount of excess oxygen left over is 36.26 grams.