If 0.5 g of a mixture of two metals A and B with respective equivalent weights 12 and 9 displace 560 ml of H2 at STP from an acid , the composition of the mixture is
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Molar gas volume at STP = 22400 mL
Moles of H2 = 560/22400
= 0.025
Mass of H2 = 0.025 x 2 = 0.05g
Let mass of metal A = x
Mass of metal B = 0.5 - x
x/12 + (0.5 - x) /9 = 0.05
Giving:
x = 0.2 (see the attached solution)
0.5 - x = 0.5 - 0.2 = 0.3
Mass of metal A = 0.2g
Mass of metal B = 0.3g
Moles of H2 = 560/22400
= 0.025
Mass of H2 = 0.025 x 2 = 0.05g
Let mass of metal A = x
Mass of metal B = 0.5 - x
x/12 + (0.5 - x) /9 = 0.05
Giving:
x = 0.2 (see the attached solution)
0.5 - x = 0.5 - 0.2 = 0.3
Mass of metal A = 0.2g
Mass of metal B = 0.3g
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