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2 years ago

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Which expression represents the percent by mass of nitrogen in NH4NO3?

2 answers:

guajiro [1.7K]2 years ago

8 0

Answer:

This question is incomplete. The completed question is below

Which expression represents the percent by mass of nitrogen in NH4NO3?

A) 14 g N/80 g NH₄ NO₃ ₓ 100%

B) 28 g N/80 g NH₄ NO₃ ₓ 100%

C) 80 g NH₄ NO₃ /14 g N ₓ 100%

D) 80 g NH₄ NO₃ /28 g N ₓ 100%

The correct option is B

Explanation:

To calculate the percentage by mass of nitrogen in the compound, the formula will be

total mass of nitrogen/total mass of compound × 100

Atomic mass of nitrogen (N) is 14 g/mol

There are 2 nitrogen atoms present in the compound as highlighted below

<u><em>Total mass of nitrogen</em></u> (N) present in NH₄ NO₃ = 14 × 2 = 28g

Atomic mass of hydrogen (H) is 1 g/mol. There are 4 hydrogen atoms present in the compound, NH₄NO₃.

Atomic mass of oxygen (O) is 16 g/mol. There are 3 oxygen atoms present in the compound, NH₄NO₃

<u><em>Total mass of the compound</em></u> =

(14 × 2) + (1 ×4) + (16 × 3) = 80g

Hence, the percentage by mass of nitrogen in the compound (as shown in the formula earlier) is

28g of N/80g of NH₄NO₃ × 100%

podryga [215]2 years ago

7 0

W(N)={2M(N)/M(NH₄NO₃)}*100%

w(N)={2*14.0 g/mol/80.0 g/mol}*100% = 35%

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As we are given that,

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Answer:

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Explanation:

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1) Balance the chemical equation.

$3Ca(NO_3)_2+2Li_3PO_4\rightarrow6LiNO_3+Ca_3(PO_4)_2$

2) List the known and unknown quantities.

Reactant 1: Ca(NO3)2.

Amount of substance: 3.40 mol.

Reactant 2: Li3PO4.

Amount of substance: 2.40 mol.

Product: Ca3(PO4)2

Mass: unknown.

3) Which is the limiting reactant?

<em>3.1-How many moles of Li3PO4 do we need to use all of the Ca(NO3)2?</em>

The molar ratio between Li3PO4 and Ca(NO3)2 is 2 mol Li3PO4: 3 mol Ca(NO3)2.

$mol\text{ }Li_3PO_4=3.40\text{ }mol\text{ }Ca(NO_3)_2*\frac{2\text{ }mol\text{ }Li_3PO_4}{3\text{ }mol\text{ }Ca(NO_3)_2}=2.2667\text{ }mol\text{ }Li_3PO_4$

<em>We need 2.2667 mol Li3PO4 and we have 2.40 mol Li3PO4. We have enough Li3PO4. </em>This is the excess reactant.

<em>3.2-How many moles of Ca(NO3)2 do we need to use all of the Li3PO4?</em>

The molar ratio between Li3PO4 and Ca(NO3)2 is 2 mol Li3PO4: 3 mol Ca(NO3)2.

$mol\text{ }Ca(NO_3)_2=2.40\text{ }mol\text{ }Li_3PO_4*\frac{3\text{ }mol\text{ }Ca(NO_3)_2}{2\text{ }mol\text{ }Li_3PO_4}=3.60\text{ }mol\text{ }Ca(NO_3)_2$

<em>We need 3.60 mol Ca(NO3)2 and we have 3.40 mol Ca(NO3)2. We do not have enough Ca(NO3)2. </em>This is the limiting reactant.

4) Moles of Ca3(PO4)2 produced from the limiting reactant.

We have 3.40 mol Ca(NO3)2 of the limiting reactant.

The molar ratio between Ca(NO3)2 and Ca3(PO4)2 is 3 mol Ca(NO3)2: 1 mol Ca3(PO4)2.

$mol\text{ }Ca_3(PO_4)_2=3.40\text{ }mol\text{ }Ca(NO_3)_2*\frac{1\text{ }mol\text{ }Ca_3(PO_4)_2}{3\text{ }mol\text{ }Ca(NO_3)_2}=1.1313\text{ }mol\text{ }Ca_3(PO_4)_2$

5) Mass of Ca3(PO4)2 produced.

The molar mass of Ca3(PO4)2 is 310.1767 g/mol.

$g\text{ }Ca_3(PO_4)_2=1.1333\text{ }mol\text{ }Ca_3(PO_4)_2*\frac{310.1767\text{ }g\text{ }Ca_3(PO_4)_2}{1\text{ }mol\text{ }Ca_3(PO_4)_2}$ $g\text{ }Ca_3(PO_4)_2=351.526\text{ }g\text{ }Ca_3(PO_4)_2$

<em>The mass of Ca3(PO4)2 produced is</em> 351 g Ca3(PO4)2.

Option D.

.

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