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@ Experiment not given. ② here loog Ae & 25mL 2.0m KoH given. 240 +2ko4 + 6tho - KAl(OH)4 +3th (a) Mass of Al available = 1.01g molar mass 2 27g/mol mole of Al = 1.01 a 0.0374you 2698 2 0.0374you =26.98glnic Now audding to storichesnetic equation above 2 mou af te need amol kou to realt.
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So 0.0374 mole AC will need 10.0374 mol Roul' 10.0374 mol kou to react 1000 a (6) How moles of kon available = volume X molarity = 25+2 0,05 mol 1000 0.05 mol of kou are available and 0.0374 mole will be consumed in the reaction to realt with Al Remaining more of fona 0.05-0.0374 = 10:0126 mole Rou (① when reaction is completed the role of Fou 20.0126 mole and volume a 25mL So molants after reactions moles x 1000 molano q ue volume amb 0.0126 X1000 25 = em 0.504 M 1 ③ Here compound a K AL(S04) 1420 molar man = 39+27 +2x(32+64 +12(16+2) = 42 Here in imolecule 12 molecule water attached to mass of water attached a 124(1642) 2 2 Watermals xlov 216 4100 mar% of water a we molar mal 474 145.
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