Question

Consider the unbalanced chemical equation below. CaSiO3(-) + HF (@)-CaF2(aq) + SiF4(g) + H2O(1) Suppose a 34.3 g sample of CaSiO3 is reacted with 31.3 L of HF at 27.0 °C and 1,00 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 Submit Hide Hints

Answer 1

The given reaction was follows: - Casioz (s) + Hf Ig) (a fe laq) + Sif418) + H₂oll) a It is given that 34.3 9 of Casio, rearts with 31.3L of Hf at 27.0°C and 1:00 am pressure a. By using PV = nRT ; we can calculate number of moles of hf. lath X 31.34 = nxo: 0821 aby k mot x 300K oona 31.3 x1 = 31.3 = 1.271 males. 0108 21 X 300 24.63 o Mass of nf = roles of nf & Molar mass of hf Mass of wf = 1.27 1 mol x 20,01 g/mol = 25.43 g Balanced reaction is as follows: - Ca si 0₃ 15) + 6Hf (g) - (af laq) + sif419) +3120 (1) According to equation :- Imole of Casio, reacts with 6 moles of HE to give I mole of late, Imole of sify &3 moles of Ho Calculating or finding the limiting reactant. 6 20101 ghorf not reacts with Imol x 17 2.24 g/mol of Ca sio en Ig of he reacts with 172.24 6x2oi01 g es 25,43 g of it will react with 172:24, 25,43 g of Ca si 03 6820101 36.499 but only 34.39 of casio, is given o la sio z is limiting reactant I male x 172.249/mol of Casio, prodeus I mal x 104.0791 g/ml of sify That is 172.24g of Casio, produus 104,07919 of sify . 34.2ge of lasio, floderes lot of a 1 x 34.3 - (20.739 of Sify/ Similary from equation it is sun that 172:24g of Casio, produces 54g of H₂o 6. 34.39 of Casio, produus sy 24 x 34.3 g of 120 = (10.759 I Mass of Sify= 20.73 g 1 of thor Mass of 1,0 = 10.75g