CaSiO3+HF = CaF2+SiF4 +H2O --( unbalance equation)
Balance equation: CaSiO3+6HF = CaF2+SiF4 +3H2O
Suppose a 34.g sample of CaSiO3 is reacted with 30.9 L of HF at 27 degree C and 1 atm. assuming the reaction goes to completion, calculate the mass (in grams) of the SiF4 and H2O produced in the reaction.
Mass SiF4=?
Mass H2O= ?
I have found the mole of CaSiO3 (.2944mol) and HF (1.255mol), but I need help to find the moles for the SiF and H2O and the ration of SiF and H2O is 1/6. Please help.
of moles in SiF4 and H2O under a chemical equation?
2011-12-15 12:34 am
回答 (1)
2011-12-15 10:13 pm
✔ 最佳答案
CaSiO3 + 6 HF → CaF2 + SiF4 + 3 H2O(34 g CaSiO3) / (116.1622 g CaSiO3/mol) = 0.29269 mol CaSiO3
(30.9 L HF) x (273 / (27 + 273)) / (22.414 L/mol) = 1.2545 mol HF
0.29269 mole of CaSiO3 would react completely with 0.29269 x (6/1) = 1.7561 mol of HF, but there is not that much HF present, so HF is the limiting reactant.
(1.2545 mol HF) x (1/6) x (104.0791 g SiF4/mol) = 22 g SiF4
(1.2545 mol HF) x (3/6) x (18.0153 g H2O/mol) = 11 g H2O
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