Balance the equation: 2Na + S --> Na2S
Using the given amount of the reactants in the reaction, calculate the amount of the product:
45.3g Na x (1 mol/22.99 g)= 1.97 mol of Na
105f S x (1 mol/ 32.06g) = 3.28 mol of S
The limiting reactant would be Na:
<span>1.97 mol Na x (1 mol Na2S/ 2 mol Na) x (78.04g/mol) = 76.87g of Na2S produced</span>
Can you post a picture of that the checkings are
The balanced equation of the reaction is:
O3(g) + NO (g) → O2 (g) + NO2 (g)
Then the ratios of reaction is 1 mol O3 : 1 mol NO : 1 mol O2 : 1 mol NO2
If you have initially 0.05 M of O3 and 0.02 M of NO, the reaction will end when all the NO is consumed.
The by the stoichiometry 0.02 mol of O3 will be consumed in 8 seconds.
And the rate of reaction is change in concetration divided by the time.
The change in concentration in O3 is 0.02 M
Then, the rate respect O3 is 0.02 M / 8 seconds = 0.0025 M/s
Electrical energy is the energy of electrons
