Question

Use the References to access important values Il meeueu lur luis question. For the following reaction, 14.3 grams of silicon tetrafluoride are allowed to react with 3.30 grams of water. silicon tetrafluoride(s) + water(1) —→ hydrofluoric acid(aq) + silicon dioxide(s) What is the maximum mass of hydrofluoric acid that can be formed? What is the FORMULA for the limiting reagent? What mass of the excess reagent remains after the reaction is complete?

Answer 1

sifu = 14:3gm H₂O = 3.30gm MM = 28.09+4(19) MM = 2(1)+ 16 =) 18 gru mo) = 104.09 ommel nH2O = 3.30 mol 16 nsifu = 14.3 mol 104.09. MMHE - 1+19 - 20gmmel Sifu (s) + 2 H₂O W UHF + SIO2 (s) initial (moles 14.3 3.30 104.09 16 Limiting 1413 > 4x330 Reagent Reogent 104.09 104.09 mor-mole analysis it will act as ciniting Reagent NH2O = THE =) 3:30/16 = nHE => nHF = 0.4125 moles > WHE = 6.0125 x 20) gm WHF = 8:25gm Limiting reagent water CH2O) Lave. » nsifu = nH2O = nsifu (used) = 3.30/16 Wsifu (wa) = ( 3:30 x 104.09) 7m 3) Wsifu cura] 10-739m Wsi fucremained] =(14.3 -10-73)ym → 3.579m Wsifu cremained = 2.57 gm / Are 2 2 the