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Sagot :
35 g al(s) reacts with excess hcl(aq) according to the chemical equation shown above. what is the volume (in l) of h2 gas produced at a temperature of 345 k and a pressure of 1.12 atm?
2Al(s) + 6HCl(aq) ----------------> 2AlCl3(aq) + 3H2(g)
3 moles of H2 produced from 2 moles of Al
3*2g of H2 produced from 2*27g of Al
75g of H2 produced from = 2*27*75/(3*2) = 675g of Al >>>>answer
2Al(s) + 6HCl(aq) ----------------> 2AlCl3(aq) + 3H2(g)
2 moles of Al react with 6 moles of HCl
2*27g of Al react with 6 moles of HCl
4.25g of Al react with = 6mole*4.25g/(2*27g) = 0.472 moles of HCl
no of moles of HCl = molarity * volume in L
0.472 = 5*volume in L
volume in L = 0.472/5 = 0.0944 L = 94.4ml >>>>answer
2Al(s) + 6HCl(aq) ----------------> 2AlCl3(aq) + 3H2(g)
2 moles of Al react with excess of HCl 3 moles of H2
2*27g of Al react with excess of HCl 3 moles of H2
35g of Al react with excess of HCl = 3*35/(2*27) = 1.94 moles of H2
n = 1.94moles
T = 345K
P = 1.12atm
PV = nRT
V = nRT/P
= 1.94*0.0821*345/(1.12) = 49.1 L of H2 >>>>answer
part-D
no of moles of Al = W/G.M.Wt
= 125/27 = 4.63moles
no of moles of HCl = molarity *volume in L
= 3.2*2.5 = 8moles
2Al(s) + 6HCl(aq) ----------------> 2AlCl3(aq) + 3H2(g)
2 moles of Al react with 6 moles of HCl
4.63 moles of Al react with = 6*4.63/2 = 13.89 moles of HCl is required
HCl is limiting reactant
6 moles of HCl react with excess of Al to gives 3 moles of H2
8 moles of HCl react with excess of Al to gives = 3*8/6 = 4 moles of H2
theoretical yield of H2 = no of moles * gram mola rmass
= 4*2 = 8g
A chemical is any substance that has a described composition. In other phrases, a chemical is continually made up of the identical "stuff." a few chemical compounds occur in nature, such as water.
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