Answer
Balanced equation:
$$2 \ AB + \ A_2 \longrightarrow 2 \ A_2B$$
If 2.50 moles of $A_2$ reacts with excess AB, 5.00 moles of $A_2B$ are produced.
The a.m.u. of $A_2B$ is 50.0 u
15.0 g of AB needs 10.0 g of $A_2$ and produces 25.0 g of $A_2B$
Work Step by Step
Balanced equation:
1. Count the molecules in each part.
Left: 8 AB and 4 $A_2$
Right: 8 $A_2B$
2. Write as an equation, putting the numbers we counted:
$$8 \ AB + 4 \ A_2 \longrightarrow 8 \ A_2B$$
3. Simplify to the smallest integers:
$$2 \ AB + \ A_2 \longrightarrow 2 \ A_2B$$
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3. Each mole of $A_2$ produces 2 moles of $A_2B$:
$$ 2.50 \space moles \space A_2 \times \frac{ 2 \space moles \ A_2B }{ 1 \space mole \space A_2 } = 5.00 \space moles \space A_2B $$
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4. If $A_2$ mass is 40.0 u, the mass of A is half of that, 20.0 u.
Therefore, the mass of B is 30.0 u (the mass of AB) minus 20.0 u, 10.0 u.
The mass of $A_2B$ is: 20.0 u * 2 + 10.0 u = 50.0 u
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5.
$$ 15.0 \space g \space AB \times \frac{1 \space mole \space AB }{ 30.0 \space g \space AB } \times \frac{ 1 \space mole \space A_2 }{ 2 \space moles \space AB } \times \frac{ 40.0 \space g \space A_2 }{1 \space mole \space A_2 } = 10.0 \space g \space A_2 $$
6.$$ 15.0 \space g \space AB \times \frac{1 \space mole \space AB }{ 30.0 \space g \space AB } \times \frac{ 2 \space moles \space A_2B }{ 2 \space moles \space AB } \times \frac{ 50.0 \space g \space A_2B }{1 \space mole \space A_2B } = 25.0 \space g \space A_2B $$
or, since we only have one product:
Mass of product = 15.0 g + 10.0 g = 25.0 g
