2na2o2 + 2co2 → 2na2co3 + o2
the percent yield of na2co3 when 7.80 g of na2o2 is reacted with excess co2 is 81.0 %.
what is the actual yield of na2co3?

if you know understand how this process works, let me know how to do it, and the answer.
you.

8.58 g.

Explanation:

To get the actual yield of a reaction, we can use the relation:

The percent yield = [actual yield]/[theoretical yield] x 100.

The percent yield = 81.0 %.

We need to calculate the theoretical yield:From the balanced reaction: 2Na₂O₂ + 2CO₂ → 2Na₂CO₃ + O₂.It is clear that 2.0 moles of Na₂O₂ react with 2.0 moles of CO₂ to produce 2.0 moles of Na₂CO₃ and 1.0 mole of O₂.The no. of moles of reacted 7.80 g of Na₂O₂ = mass/molar mass = (7.80 g)/(77.98 g/mol) = 0.1 mol.We can get the no. of moles of produced Na₂CO₃ using cross multiplication:

∵ 2.0 moles of Na₂O₂ produce → 2.0 moles Na₂CO₃.

∴ 0.1 mol of Na₂O₂ produce → 0.1 mol Na₂CO₃.

We can get the mass of 0.1 mol Na₂CO₃:

mass of 0.1 mol Na₂CO₃ = n x molar mass = (0.1 mol)(105.98 g/mol) = 10.598 g.

∵ The percent yield = [actual yield]/[theoretical yield] x 100.

∴ 81.0 % = [actual yield]/[10.598 g] x 100.

∴ [actual yield] = (81.0 %)(10.598)/(100) = 8.58 g.

It was ernest rutherford

Ibelieve it's celsius. hope this .