In a) the final equation is AgNO3 + KCl = AgCl + KNO3, b) Ni(NO3)2 + Na2S = 2NaNO3 + NiS; c) CaCl2 + Na2CO3 = 2 NaCl + CaCO3. In 2) The total net equation is Ca 2+ + CO32- = CaCO3 (s).
Answer:
n(HCl)=1.96 mol
Explanation:
CH4+4Cl2⟶CCl4+4HCl
CCl4+2HF⟶CCl2F2+2HCl
With ideal yields we will end up with 4 moles of HCl.
With 70% yields on every stage
n(HCl)=0.7*0.7*4=1.96 mol
0.92%. student calculated that the amount of NaCl that should form in the lab
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