Question

please show work

How many grams of HCl, are needed to produce 125 ml of carbon dioxide gas at STP in the following reaction? CaCO3(s) + 2 HCl(aq) - -> CaCl(aq) + CO2(g) + H20(1)

Answer 1

16 CaCO3(s) + 2HClap, CaCl₂ cap. st CO₂(g) + Ou) Given Vecm) = 125m) at STP Imole = 22.400m) o gus moles) (0, Obtaimel - 125 mi mol 22400 m By Reachim = 0.00558 mol & 2 mole of Hel will Produce 1 mole of CO₂ moles) HCl equired = 2X 0.00558 mot = 0. 01/16 mol mand. HCL = moles X mokur man o HCL = 0.011168 (35.5+ 1) c1=3559/not = 0.40 gram An. H 1 9Imo ③ 2 A1+ Ghcl 2 AlCl3 + 3H₂ man 0.625 0.625 Moles of Al = may - 0.02316 mol - Molarmen de AL 26.9815 2 moles of Al produces 3 mole of H₂ = 0.02316 mole o Al Will Produce = 3x0:02316 mole Hz 260 = 0.03474 mol Given = P = 759 mm of Hg - =769 atm मि- (ana = 0.99868 atm He measure (data mining) T= 55.0°C = 328K By ideal gas equation e- P PV=MRT" V= 0.03474 X0.0821 X 328 - 0.936 L 0.99868 Prenure is giren in Table

Macs) tancl - Mg ch t h may 265 kg. man of Mq ing E 261 X Low 5265.000 galams molaut mam of My = 24.3 gram '- moles o mg = 10.905.349 miles By Balanced Reaction, (mole of Macs) will produce a mole d H at STP =) Same moles of Hy will be produced moles of H₂ 9 = 10,905. 349 Imole of gas = 22.4 Litres at STP a volume of H (9) at STP = moles X 22.4. = 10,905. 3498 22.4 L = 244279.81 litres of th (9) Do upole