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Radda [10]
3 years ago
6

How many grams of fluorine are needed to react with 8.5g of phosphorus?

Chemistry
1 answer:
love history [14]3 years ago
4 0

Step 1: write the equation:

                                          P₄(s) + 6F₂(g) → 4PF₃(g) 


Step 2: Molar mass of P₄ = 30.97 g/mol × 4 = 123.88 g/mol


Step 3: Number of moles of phosphorus

                                                    n = m/M 

                                                    n = 8.5 g/123.88g/mol

                                                    n = 0.07 moles


Step 4: 0.07 × 12 = 0.84 moles of fluorine.

Fluorine is diatomic gas so we multiplied the number of moles by 12.


Step 5: To find the mass of fluorine we multiply the number of moles with the molar mass.

                            Mass of fluorine = 0.84 ×  228 

                                                        = 191.52 grams.

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Find the percent composition of C7H5N3O6
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Finding percent composition is fairly easy. You only need to divide the mass of an element by the total mass of the compound. We can do this one element at a time.

First, let's find the total mass by using the masses of the elements given on the periodic table.

7 x 12.011 (mass of Carbon) = 84.077

5 x 1.008 (mass of Hydrogen) = 5.04

3 x 14.007 (mass of Nitrogen) = 42.021

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Add all of those pieces together.

84.077 + 5.04 + 42.021 + 95.994 = 227.132 g/mol is your total. Since we also just found the mass of each individual element, the next step will be very easy.

Carbon: 84.077 / 227.132 = 0.37016 ≈ 37.01 %

Hydrogen: 5.04 / 227.132 = 0.022189 ≈ 2.22 %

Nitrogen: 42.021 / 227.132 = 0.185 ≈ 18.5 %

Oxygen: 95.994 / 227.132 = 0.42263 ≈ 42.26 %

You can check your work by making sure they add up to 100%. The ones I just found add up to 99.99, which is close enough. A small difference (no more than 0.03 in my experience) is just a matter of where you rounded your numbers.

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