The limiting reactant is the substance of the reactants that runs out first. The excess reactant is the substance with the greater amount than necessary to react completely with the limiting reactant after a chemical reaction.
Well you use the atomic mass of helium and multiply by the 3.75 grams. Then you will get the answer around 3.7875
Answer:
There are 6.024 grams of sodium hydroxide in the solution.
Explanation:
![Molarity=\frac{Moles}{Volume (L)}](https://tex.z-dn.net/?f=Molarity%3D%5Cfrac%7BMoles%7D%7BVolume%20%28L%29%7D)
![Moles (n)=Molarity(M)\times Volume (L)](https://tex.z-dn.net/?f=Moles%20%28n%29%3DMolarity%28M%29%5Ctimes%20Volume%20%28L%29)
Moles of sodium hydroxide = n
Volume of sodium hydroxide solution = 251.0 mL = 0.251 L
Molarity of the sodium hydroxide = 0.600 M
![n=0.600 M\times 0.251 L=0.1506 mol](https://tex.z-dn.net/?f=n%3D0.600%20M%5Ctimes%200.251%20L%3D0.1506%20mol)
Mass of 0.1506 moles of NaOH :
![40 g/mol\times 0.1506 mol = 6.024 g](https://tex.z-dn.net/?f=40%20g%2Fmol%5Ctimes%200.1506%20mol%20%3D%206.024%20g)
There are 6.024 grams of sodium hydroxide in the solution.
Answer:
Barium carbonate powder is stirred add pulp in the entry, the vitriol that the adds solubility then reaction that makes the transition is filtered and is obtained the barium sulfate filter cake and liquid after the transition.