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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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How many liters of hydrogen gas will be produced at STP from the reaction of 7.179 x 1023 atoms of magnesium with 54.21 g of pho
wlad13 [49]
<h3>Answer:</h3>

18.58 liters of hydrogen gas

<h3>Explanation:</h3>

We are given;

  • The equation;

3Mg + 2H₃(PO₄) → Mg₃(PO₄)₂ + 3H₂

  • Atoms of Magnesium = 7.179 x 10^23 atoms
  • Mass of phosphoric acid as 54.21 g

We are required to determine the volume of hydrogen gas produced;

Step 1; moles of Magnesium

1 mole of an element contains 6.02 × 10^23 atoms

therefore;

Moles of Mg = (7.179 x 10^23 ) ÷ (6.02 × 10^23)

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Step 2: Moles of phosphoric acid

moles = Mass ÷ Molar mass

Molar mass of phosphoric acid = 97.994 g/mol

Therefore;

Moles of Phosphoric acid = 54.21 g ÷ 97.994 g/mol

                                           = 0.553 moles

Step 3: Determine the rate limiting reagent

From the mole ratio of Mg to Phosphoric acid (3 : 2);

1.193 moles of magnesium requires 0.795 moles of phosphoric acid while,

0.0553 moles of phosphoric acid requires 0.8295 moles of Mg

Therefore, phosphoric acid is the rate limiting reagent

step 4: Determine the moles of hydrogen produced

From the equation, w moles of phosphoric acid reacts to produce 3 moles of hydrogen;

Therefore; moles of Hydrogen = moles of phosphoric acid × 3/2

                                                   = 0.553 moles × 3/2

                                                   = 0.8295 moles

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1 mole of a gas occupies a volume of 22.4 liters at STP

Therefore;

Volume of Hydrogen = 0.8295 moles × 22.4 L/mol

                                  = 18.58 Liters

Therefore; 18.58 liters of hydrogen gas  will be produced

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