Given the balanced equation:
( Reaction type : double replacement)
CaF2 + H2SO4 → CaSO4 + 2HFI
We can determine the number of grams prepared from the quantity of 75.0 H2SO4, and 63.0g of CaF2 by converting these grams to moles per substance.
This can be done by evaluating the atomic mass of each element of the substance, and totaling it to find the molecular mass.
For H2SO4 or hydrogen sulfate it's molecular mass is the sum of the quantity of atomic mass per element. H×2 + S×1 + O×4 = ≈1.01×2 + ≈32.06×1 + ≈16×4 = 2.02 + 32.06 + 64 = 98.08 u (Dalton's or Da) or g / mol.
For CaF2 or calcium fluoride, it's molecular mass adds 1 atomic mass of calcium and 2 atomic masses of fluoride due to the number of atoms.
Ca×1 + F×2 = ≈40.07×1 + ≈19×2 = 40.08 + 38 = 78.07 u (Da or Dalton's) or g / mol.
It is B, and also for a moment I didn't understand that 4.69 x 10^22. I almost did this whole problem wrong.
Balance the equation first:
2 Fe+6 HNO3→2 Fe(NO3)3+3H2
Then calculate mass of Iron :
4.5×3.0×3.5 cm3(1 mL1 cm3)(7.87 g Fe1 ml)=371.86 g Fe
Now use Stoichiometry:
371.86 g Fe×(1 mol Fe55.85 g Fe)×(6 mol HNO32 mol Fe)=19.97 mol HNO3
Convert moles of nitric acid to grams
19.97 mol HNO3×(63.01 g HNO31 mol HNO3)=1258.3 g HNO3
I think C. Mutualism.
Hope this helps :)
