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2Br +7 02 -> 2 Br2 07

If 425 grams of oxygen gas reacted with 350 grams of bromine, what should be the theoretical yield of dibromine heptoxide? chart sample ​

User Pree
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Answer:

The balanced chemical equation is:

2Br + 7O2 → 2Br2O7

From the equation, we can see that 7 moles of oxygen are required to react with 2 moles of bromine to form 2 moles of dibromine heptoxide.

First, we need to determine the limiting reagent:

The moles of oxygen gas can be calculated as:

n(O2) = m/M = 425 g / 32 g/mol = 13.28 mol

The moles of bromine can be calculated as:

n(Br) = m/M = 350 g / 80 g/mol = 4.375 mol

The limiting reagent is bromine because it is present in a smaller quantity.

Theoretical yield of dibromine heptoxide can be calculated using the moles of limiting reagent:

n(Br2O7) = n(Br) × (2 mol Br2O7 / 2 mol Br) × (7 mol O2 / 2 mol Br2O7) = 4.375 × 7 / 2 = 15.3125 mol

The molar mass of Br2O7 is:

M(Br2O7) = 2 × M(Br) + 7 × M(O) = 2 × 80 g/mol + 7 × 16 g/mol = 296 g/mol

The theoretical yield of dibromine heptoxide can be calculated as:

mass(Br2O7) = n(Br2O7) × M(Br2O7) = 15.3125 mol × 296 g/mol = 4534.375 g or 4.53 kg

Therefore, the theoretical yield of dibromine heptoxide is 4.53 kg.

User Damon Baker
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