Yes. carbon dioxide is deadly and i want to get away from it fast.
Answer: 51.4 g of NaClO will be produced.
Explanation:
To calculate the moles :
![\text{Moles of} Cl_2=\frac{48.9g}{71g/mol}=0.69moles](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%7D%20Cl_2%3D%5Cfrac%7B48.9g%7D%7B71g%2Fmol%7D%3D0.69moles)
![\text{Moles of} NaOH=\frac{54.2g}{40g/mol}=1.4moles](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%7D%20NaOH%3D%5Cfrac%7B54.2g%7D%7B40g%2Fmol%7D%3D1.4moles)
According to stoichiometry :
1 moles of
require = 2 moles of ![NaOH](https://tex.z-dn.net/?f=NaOH)
Thus 0.69 moles of
will require=
of ![NaOH](https://tex.z-dn.net/?f=NaOH)
Thus
is the limiting reagent as it limits the formation of product and
is the excess reagent.
As 1 mole of
give = 1 mole of ![NaClO](https://tex.z-dn.net/?f=NaClO)
Thus 0.69 moles of
give =
of ![NaClO](https://tex.z-dn.net/?f=NaClO)
Mass of ![NaClO=moles\times {\text {Molar mass}}=0.69moles\times 74.5g/mol=51.4g](https://tex.z-dn.net/?f=NaClO%3Dmoles%5Ctimes%20%7B%5Ctext%20%7BMolar%20mass%7D%7D%3D0.69moles%5Ctimes%2074.5g%2Fmol%3D51.4g)
Thus 51.4 g of NaClO will be produced
Na, Ba, Cs and Ni are metals.
Answer: The answer is D.gram
Explanation:
Answer:
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Explanation: