50. 3 g of an alloy of aluminium and magnesium wastreated with dilute

1.	50. 3 g of an alloy of aluminium and magnesium wastreated with dilute HCl (excess). Aluminium andmagnesium react with HCl as per followingreaction :2AI + 6HCI 2AlCl3 + 3H2Mg + 2HCl → MgCl2 + H2The hydrogen produced has a volume of 3.136 Lat STP. The percentage composition ofaluminium by mass in the alloy is (Mg = 24 g/mol,Al = 27 g/mol)
Answer» Answer: hey, it should be 39 g of ALLOY! Explanation: let the MOLE of MAGNESIUM be x and mole of aluminum be y Mg + 2Hcl---MgCL2 +H2 Al +3Hcl ----AlCl3 +1.5H2 NOW 24x +27y=39 x +1.5y=2 solving these two equation we get x=.5 ,y=1 percentage of magnesium=30.76% percentage of aluminum=69.24% APPLYING PV=nRT for hydrogen n=2 mole let the mole of magnesium be x and mole of aluminum be y Mg + 2Hcl---MgCL2 +H2 Al +3Hcl ----AlCl3 +1.5H2 NOW 24x +27y=39 x +1.5y=2 *solving these two equation we get x=.5 ,y=1 percentage of magnesium=30.76% percentage of aluminium=69.24%*

50. 3 g of an alloy of aluminium and magnesium wastreated with dilute HCl (excess). Aluminium andmagnesium react with HCl as per followingreaction :2AI + 6HCI 2AlCl3 + 3H2Mg + 2HCl → MgCl2 + H2The hydrogen produced has a volume of 3.136 Lat STP. The percentage composition ofaluminium by mass in the alloy is (Mg = 24 g/mol,Al = 27 g/mol)

Answer»

Answer:

hey, it should be 39 g of ALLOY!

Explanation:

let the MOLE of MAGNESIUM be x and mole of aluminum be y

Mg + 2Hcl---MgCL2 +H2

Al +3Hcl ----AlCl3 +1.5H2

NOW

24x +27y=39

x +1.5y=2

solving these two equation we get

x=.5 ,y=1

percentage of magnesium=30.76%

percentage of aluminum=69.24%

APPLYING PV=nRT for hydrogen

n=2 mole

let the mole of magnesium be x and mole of aluminum be y

Mg + 2Hcl---MgCL2 +H2

Al +3Hcl ----AlCl3 +1.5H2

NOW

24x +27y=39

x +1.5y=2

solving these two equation we get x=.5 ,y=1 percentage of magnesium=30.76% percentage of aluminium=69.24%