To find out how many grams of reactant A are needed to produce 112 g of the product A2B, we can follow these steps:
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Determine the molar mass of A2B: Already provided as 28 g/mol.
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Calculate the number of moles of A2B in 112 g: \[ \text{moles of A2B} = \frac{\text{mass of A2B}}{\text{molar mass of A2B}} = \frac{112 , \text{g}}{28 , \text{g/mol}} = 4 , \text{moles} \]
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Use the balanced chemical equation to find the ratio of A to A2B: The equation is: \[ 4A + B2 \rightarrow 2A2B \] From this equation, we see that 4 moles of A produce 2 moles of A2B. Therefore, for every 2 moles of A2B produced, 4 moles of A are required.
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Calculate how many moles of A are needed for 4 moles of A2B: From the stoichiometry, if 4 moles of A2B require 8 moles of A (since \( \frac{4 \text{ moles of A}}{2 \text{ moles of A2B}} = 2 \) times for 4 moles of A2B): \[ \text{moles of A needed} = 8 \text{ moles} \]
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Convert moles of A needed to grams: \[ \text{mass of A} = \text{moles of A} \times \text{molar mass of A} = 8 , \text{moles} \times 12 , \text{g/mol} = 96 , \text{g} \]
So, the initial amount of reactant A needed is 96 grams.
Final Answer: 96