A Gas Stoichiometry Liters of a Gas STP

A. Gas Stoichiometry Liters of a Gas • STP - use 22. 4 L/mol • Non-STP - use ideal gas law b Moles b Non-STP Problems • Given liters of gas? Ø start with ideal gas law • Looking for liters of gas? Ø start with stoichiometry conv.

B. Gas Stoichiometry Problem b What volume of CO 2 forms from 5. 25 g of Ca. CO 3 at 103 k. Pa & 25ºC? Ca. CO 3 5. 25 g Ca. O Looking for liters: Start with stoich and calculate moles of CO 2. 5. 25 g 1 mol Ca. CO 3 CO 2 + CO 2 ? L non. STP = 1. 26 mol 100. 09 1 mol Plug. CO this into 2 the Ideal g Ca. C Gas Law to find liters.

B. Gas Stoichiometry Problem b What volume of CO 2 forms from 5. 25 g of Ca. CO 3 at 103 k. Pa & 25ºC? GIVEN: WORK: P = 103 k. Pa V=? n = 1. 26 mol T = 25°C = 298 K R = 8. 315 PV = n. RT (103 k. Pa)V =(1 mol)(8. 315 dm 3 k. Pa/mol K)( 298 K) dm 3 k. Pa/mol K V = 1. 26 dm 3 CO 2

B. Gas Stoichiometry Problem b How many grams of Al 2 O 3 are formed from 15. 0 L of O 2 at 97. 3 k. Pa & 21°C? 4 Al + GIVEN: P = 97. 3 k. Pa V = 15. 0 L n=? T = 21°C = 294 K R = 8. 315 3 O 2 15. 0 L non. STP WORK: 2 Al 2 O 3 ? g Given liters: Start with Ideal Gas Law and calculate moles of O 2. PV = n. RT (97. 3 k. Pa) (15. 0 L) = n (8. 315 dm 3 k. Pa/mol K) NEXT (294 K)

B. Gas Stoichiometry Problem b How many grams of Al 2 O 3 are formed from 15. 0 L of O 2 at 97. 3 k. Pa & 21°C? 3 O 2 Use stoich to convert moles 15. 0 L of O to grams Al O. non 0. 597 2 mol STP 101. 96 mol O 2 Al 2 O 3 g Al 2 O 3 4 Al 2 2 + 2 Al 2 O 3 ? g 3 3 mol O 2 = 40. 6 g 1 mol Al 2 O 3

C. Gas Stoichiometry Problem b Calculate the volume of O 2 at STP required for complete combustion of 7. 64 L of acetylene (C 2 H 2) at STP? 2 C 2 H 2(g) + 5 O 2(g) 4 CO 2(g) + 2 H 2 O(l) b At STP, so we can convert directly to liters! b b 7. 64 L C 2 H 2 x (5 L O 2 / 2 L C 2 H 2) = 19. 1 L