What is the theoretical yield of nacl would result from reacting 5.00 mil of na2co3 with excess hcl?

584.4 g.

Explanation:

Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂.

It is clear that 1.0 mole of Na₂CO₃ reacts with HCl to produce 2.0 moles of NaCl.

Using cross multiplication:

1.0 mole of Na₂CO₃ → 2.0 moles of NaCl.

5.0 moles of Na₂CO₃ → ??? moles of NaCl.

The no. of moles of NaCl produced from 2.0 moles of Na₂CO₃ = (2.0 mol)(5.0 mol)/(1.0 mol) = 10.0 mol.

∴ The theoretical yield of NaCl = no. of moles of NaCl x molar mass of NaCl = (10.0 mol)(58.44 g/mol) = 584.4 g.

in scientific notation, the value of the  speed of light in a vacuum

  will be written as 2.995 * 10^8. when this value is represented in the form of

  a * 10^b, then, a = 2.995 and b = 8.

the value of the speed of light in a vacuum given above can also be rounded up to approximately 300000000. in this case, the scientific notation will be written as 3 *10^8.