1. Barium chloride and sodium sulfate react according to the following equation.
ID: 1026955 • Letter: 1
Question
1. Barium chloride and sodium sulfate react according to the following equation.
BaCl2 + Na2SO4 BaSO4 + 2NaCl
How many grams of barium sulfate can be produced from 23.2 g of barium chloride?
2.How many molecules of HCl are formed when 40.8 g of water reacts according to the following balanced reaction? Assume excess ICl3.
2 ICl3 + 3 H2O ICl + HIO3 + 5 HCl
3.The density of ethanol, C2H5OH, is 0.789 g/mL. How many milliliters of ethanol are needed to produce 59.5 g of CO2 according to the following chemical equation?
C2H5OH(l) + 3 O2(g) 2 CO2(g) + 3 H2O(l)
Explanation / Answer
1. BaCl2 + Na2SO4 BaSO4 + 2NaCl
1 mol BaCl2 = 1 mol Na2SO4
no of mol of barium chloride = w/M = 23.2/208.23
= 0.111 mol
Na2SO4 = excess reagent
no of mol of barium sulfate = 0.111 mol
amount of barium sulfate = n*M
= 0.111*233.38
= 25.9 g
2.
2 ICl3 + 3 H2O ICl + HIO3 + 5 HCl
2 mol ICl3 = 3 mol H2O
NO of mol of water = 40.8/18 = 2.267 mol
No of mol of HCl formed = 2.267*5/3 = 3.78 mol
No of molecules of HCl are formed = 3.78*6.023*10^23
= 2.28*10^24 molecules
3.
C2H5OH(l) + 3 O2(g) 2 CO2(g) + 3 H2O(l)
No of mol of CO2 formed = 59.5/44 = 1.352 mol
No of mol of C2H5OH REQUIRED = 1.352*(1/2) = 0.676 mol
Amount of C2H5OH required = 0.676*46 = 31.1 g
volume of C2H5OH required = m/d
= 31.1/0.789
= 39.42 ml
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