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faust18 [17]
3 years ago
12

Assuming an efficiency of 30.80%, calculate the actual yield of magnesium nitrate formed from 147.4g of magnesium and excess cop

per(II) nitrate. Write the number in grams.
Chemistry
1 answer:
Reil [10]3 years ago
4 0

Answer:

The answer to your question is 280 g of Mg(NO₃)₂

Explanation:

Data

Efficiency = 30.80 %

Mg(NO₃)₂ = ?

Magnesium = 147.4 g

Copper (II) nitrate = excess

Balanced Reaction

                     Mg  +   Cu(NO₃)₂    ⇒    Mg(NO₃)₂   +   Cu

                  Reactants           Elements            Products

                         1                          Mg                     1

                         1                          Cu                      1

                         2                          N                       2

                         6                          O                       6

Process

1.- Calculate the theoretical yield

Molecular weight Mg = 24

Molecular weight Mg(NO₃)₂ = 24 + (14 x 2) + (16 x 6)

                                              = 24 + 28 + 96

                                              = 148 g

                           24 g of Mg  --------------------  148 g of Mg(NO₃)₂

                          147.4 g of Mg -------------------   x

                            x = (147.4 x 148) / 24

                            x = 908.96 g of Mg(NO₃)₂

2.- Calculate the Actual yield

yield percent = \frac{actual yield}{theoretical yield}

Solve for actual yield

Actual yield = Yield percent x Theoretical yield

Substitution

Actual yield = \frac{30.8}{100} x 908.96

Actual yield = 279.95 ≈ 280g

                     

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