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solniwko [45]
3 years ago
15

How many moles of N2O5 are needed to produce 7.90 g of NO2?

Chemistry
1 answer:
Roman55 [17]3 years ago
4 0

Answer:

0.085 moles of  N₂O₅ are needed

Explanation:

Given data:

Mass of NO₂ produces = 7.90 g

Moles of N₂O₅ needed = ?

Solution:

2N₂O₅       →     4NO₂  + O₂

Number of moles of NO₂ produced :

Number of moles = mass/ molar mass

Number of moles = 7.90 g/ 46 g/mol

Number of moles = 0.17 mol

now we will compare the moles of NO₂   with N₂O₅.

                NO₂          :       N₂O₅

                  4            :          2

                0.17          :         2/4×0.17 = 0.085 mol

Thus, 0.085 moles of  N₂O₅ are needed.

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Explanation:

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The formula for the heat of fusion is given by :

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Answer:  8.30 g of calcium sulfate are produced from 10 grams of lithium sulfate.

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} Ca(NO_3)_2=\frac{10g}{164g/mol}=0.061moles

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According to stoichiometry :

1 mole of Ca(NO_3)_2 require = 1 mole of Li_2SO_4

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Thus Ca(NO_3)_2 is the limiting reagent as it limits the formation of product and Li_2SO_4 is the excess reagent.

As 1 mole of Ca(NO_3)_2 give = 1 mole of CaSO_4

Thus 0.061 moles of Ca(NO_3)_2 give =\frac{1}{1}\times 0.061=0.061moles  of CaSO_4

Mass of CaSO_4=moles\times {\text {Molar mass}}=0.061moles\times 136g/mol=8.30g

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Explanation:

hope this helped you

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