Question #ec36a

1 Answer
Aug 14, 2017

Answer:

(a) 5.00 mol; (b) 50.0 u; (c) 10.0 g; (d) 25.0 g.

Explanation:

(a) Mass of #"A"_2"B"#

Step 1. Start with the balanced equation.

Your equation is not in its lowest terms. It should be

#M_text(r):color(white)(m) 40.0color(white)(ml)30.0color(white)(mml)50.0#
#color(white)(mmmll)"A"_2 + "2AB" → "2A"_2"B"#

Step 2. Calculate the moles of #"A"_2"B"#.

#"Moles of A"_2"B" = 2.50 color(red)(cancel(color(black)("mol A"_2))) × ("2 mol A"_2"B")/(1 color(red)(cancel(color(black)("mol A"_2)))) = "5.00 mol A"_2"B"#

(b) Molecular mass of #"A"_2"B"#

If #"Molec. mass of A"_2 = "40.0 u"#, then #"at. mass of A" = "20.0 u"#

#"Molec. mass of AB" = "30.0 u" = "20.0 u + at. mass of B"#

#"At. mass of B" = "30.0 u - 20.0 u" = "10.0 u"#

#"Molec. mass of A"_2"B" = "2 × 20.0 u + 10.0 u" = "40.0 u + 10.0 u" = "50.0 u"#

(c) Mass of #"A"_2# required

Step 1. Calculate the moles of #"AB"#

#"Moles of AB" = 15.0 color(red)(cancel(color(black)("g AB"))) × (1 "mol AB")/(30.0color(red)(cancel(color(black)("g AB")))) = "0.500 mol AB"#

Step 2. Calculate the moles of of #"A"_2#

#"Moles of A"_2 = 0.500 color(red)(cancel(color(black)("mol AB"))) × "1 mol A"_2/(2 color(red)(cancel(color(black)("mol AB")))) = "0.250 mol A"_2#

Step 3. Calculate the mass of #"A"_2#

#"Mass of A"_2 = 0.250 color(red)(cancel(color(black)("mol A"_2))) × "40.0 g A"_2/(1 color(red)(cancel(color(black)("mol A"_2)))) = "10.0 g A"_2#

(d) Mass of #"A"_2"B"# formed

Step 1. Calculate the moles of #"A"_2"B"# formed

#"Moles of A"_2"B" = 0.500 color(red)(cancel(color(black)("mol AB"))) × ("2 mol A"_2"B")/(2 color(red)(cancel(color(black)("mol AB")))) = "0.500 mol A"_2"B"#

Step 2. Calculate the mass of #"A"_2"B"#

#"Mass of A"_2"B" = 0.500 color(red)(cancel(color(black)("mol A"_2"B"))) × ("50.0 g A"_2"B")/(1 color(red)(cancel(color(black)("mol A"_2"B")))) = "25.0 g A"_2"B"#