In a combination reaction, 1.50 g of lithium is mixed with 6.98 g of oxygen. (a) Which reactant is present in excess? lithium. oxygen. (b) How many moles of product are formed? mol (c) After the reaction, how many grams of each reactant and product are present?

Oxygen

0,108 moles

0g Li; 5,25g of O₂; 3,23g of Li₂O

Explanation:

The reaction is:

4Li + O₂ → 2Li₂O

a) Moles of lithium in 1,50g are:

1,50g × (1mol / 6,941g) = 0,216 moles of Li

Moles of O₂ are:

6,98g × (1mol / 32g) = 0,218 moles of O₂

The complete reaction of oxygen requires:

0,218 mol ₓ (4mol Li / 1 mol O₂) = 0,872 moles of Li.

As there are more moles of oxygen than required moles of lithium the reactant in excess is oxygen

b) The limiting reactant is lithium, that means moles of product formed are moles of lithium ÷2, that is, 0,108 moles

a) After the reaction there are 0 moles of Li, 0,218 - 0,216× (1 mol O₂/4 mol Li) = 0,164 moles of O₂ and 0,432 moles of Li₂O. In grams:

0g Li

0,164 mol ₓ (32g / mol) = 5,25g of O₂

0,108mol Li₂O ₓ (29,88g / mol) = 3,23g of Li₂O

To know your answer is right you can sum the initial total mass and must be the same of final total mass

I hope it helps!

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :