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4 years ago

How many grams of Br are required to react completely with 29.5 g of NO? 2NO+Br2- 2NOBr

Chemistry

1 answer:

umka21 [38]4 years ago

8 0

Answer: 78.54 grams

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

$\text{Number of molesof} NO=\frac{29.5g}{30.01g/mol}=0.98moles$

$2NO+Br_2\rightarrow 2NOBr$

According to stoichiometry :

2 moles of $NO$ react with 1 mole of $Br_2$

0.98 moles of $NO$ react with = $\frac{1}{2}\times 0.98=0.49$ moles of $Br_2$

Mass of $Br_2=moles\times {\text{Molar mass}=0.49\times 159.8=78.54g$

Thus 78.54 g of $Br_2$ are required to react completely with 29.5 g of NO.

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The reactants are on the left side of the equation. Although, energy (heat) is a reactant it will not factor into the calculation for moles. The coefficient is the number of moles for each substances. So for the reactant side NO would have 2 moles because the coefficient is 2. Using the same logic, NO on the product side will have 2 moles and O2 will be 1 mole. O2 has one mole because it is implied that you know anything multipled by one is the same number. So 1 mole of O2 is written as O2 and not 1 O2.

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explanation
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