Question

Practice for test Al hade Think About It What is the percent composition of oxygen in sodium NA 1232 Bent composition of oxygen in sodium bromate> NATO, (1) 20 1 0- 1100 - 2 2. A certa24 for Pond is analyzed and found to have 12.0 g of C, 5g of H. and 64 g of O. What is its empirical If the molecular formula mas 495 /mol what is the molecular formala? 151 3. A hydrocarbon is to rocarbon is found to have an empleal formula of CH. I density at STP is 2.5 molecular formula? L What is its 4. Zinema Anc metal reacts with hydrochloric acid to produce hydrogen gas. If 130 grams of Zinc react, how many liters of hydrogen gas form at STPP 17001- 2001, 190 g 21 / Imo m ci 11 - 44.9L Hz 1 1u59201 molz in H Per to this BALANCED equation for questions 5-9 A + B C (5.) How many liters of C can be formed when 3 L of A react with 4L of B at STP? CLAY MdAh (22.4L0L0 9L of UW 1.) 224 LAJM ) Bis the limiting ( mBI mell/22.410-4LV Pacit. 1 7 1.91 l b ime. The molar mass of compounds 6.) Two moles of A are reacted with three mote of B. The molar mass of compound Cis 40.0 g/mol. How many grams of C can be produced during this reaction? (2 mol A WA (7.) How many liters of C can be formed when 4 L of A reacts with 5 L of B? Try to think about what kind of problem this is... Would one of the reactants limit the amount of product that can be made? (4 (LA) (MOA) ( 1 6) 22:41 4 )=40. The rooctawt A would T' 22.4LAT A (Tmoro ) Tiit Hiu niligunt of that (LD) (1 melB (1M2 /22.410-510 would wu woda pecall it 1 7 124L6 i molt i molt lomiting roaciant. 3 33

Answer 1

-X100 - 151 81 mass 1. soln : sodium bromate NaB03 %.composition of ö - mass of oxygen MW Naby03 16x3 3 *100 = 31.79% 129 C = imole carbon 59H - 5 mole H Hence, empir 646 0 = 4 mole 0 = CH504 Now empirical mass) xn = moleculai mass > sixn = 405 so, molecular formula → n: 405-5 = nxempirical > 3. At STP : density=2.5312 5,5(CH504)= C5 Has 020 density=2.58" Empirical mass in at STP imole gas = 22.44 = 22.4x2.5 = molecular -569 -> 14xn=56 Empirical formula = CH2 =*=4 mass = 14 Hence molecular formula = 4(CH2) = C4H8 4. Zn +240 EnCl2 + H2 Bog mole of Hz gas produced = 2mole - 130 - 2mole - 2x224 L volume of Mz gas "= 44.8L Ang Note: At STP Imole gas = 22.42 5. A + B => Hence, 3L of Ċ will be to 3L 4L t-final 3-3 4-7 6: A+B= +-o zmole smole je 2x400 = 809 ofc t-final 2-2 2 3 2 is produced. 7. A + B - Hence 41 of Ĉ is produced. t=0 4L SL - 3 tone formed. 2mole ofc is produced