Answer:
C the mass of each product formed
Explanation:
To the determine the limiting reactant, it is essential that we have the balanced equation of the reaction from which we can calculate the stochiometry mole ratio of the reactant. After this, we need to calculate the molar mass of the reactants, using the mole from the balanced equation we can calculate each mass of each reactant needed. Finally we need the mass of each reactant using proportion we can calculate the amount needed for the reaction from the masses of the reactant by comparing the mass given against the mass calculated from the balanced equation. After this, the mass that is exhausted or that is finished will be the limiting reactant which is the reactant that finished and caused the reaction to stop.
It should be 8 O atoms. 3O atoms in Na2S2O3 and 5O atom in 5H2O. The reason there are 5 O atoms are because the 5 in front of H2O means you multiply each atom in the compound by that number (like the distributive property). The H2 molecule becomes 10 Hydrogen atoms (5*2) and the Oxygen becomes 5 Oxygen atoms (5*1). Then you add the 5O atoms to the 3O atoms which equals 8
Use M1V1 = M2V2 to solve
3(V1) = 2.8 * 1.6
3(V1) = 4.48
V1 = 1.493 L of stock solution
Answer:
A unit of mass used to express atomic and molecular weights, equal to one-twelfth of the mass of an atom of carbon-12. It is equal to approximately 1.66 x 10-27 kg.
Explanation:
