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GaryK [48]
3 years ago
9

A chemist plans to use 435.0 grams of ammonium nitrate ion a reaction. How many moles of the compound is this?

Chemistry
1 answer:
AURORKA [14]3 years ago
4 0
Ammonium Nitrate has the formula NH4NO3, and has a formula weight of 80.043 grams/mole. We can convert the amount in the problem to moles using this conversion factor. Divide 435 by 80.043 to find the number of moles.

435/80.043 = 5.435

This amount of Ammonium Nitrate contains 5.44 moles of substance, adjusted for significant figures.
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5 0
4 years ago
The following reaction shows the products when sulfuric acid and aluminum hydroxide react.
Elden [556K]

The correct answer is approximately 11.73 grams of sulfuric acid.

The theoretical yield of water from Al(OH)3 is lower than that of H₂SO₄. As a consequence, Al(OH)3 is the limiting reactant, H₂SO₄ is in excess.

The balanced equation is:

2Al(OH)₃ + 3H₂SO₄ ⇒ Al₂(SO₄)₃ + 6H₂O

Each mole of Al(OH)3 corresponds to 3/2 moles of H₂SO₄. The molecular mass of Al(OH)3 is 78.003 g/mol. There are 15/78.003 = 0.19230 moles of Al(OH)3 in the five grams of Al(OH)3 available. Al(OH)3 is in limiting, which means that all 0.19230 moles will be consumed. Accordingly, 0.19230 × 3/2 = 0.28845 moles of H₂SO₄ will be consumed.

The molar mass of H₂SO₄ is 98.706 g/mol. The mass of 0.28845 moles of H₂SO₄ is 0.28845 × 98.706 = 28.289 g

40 grams of sulfuric acid is available, out of which 28.289 grams is consumed. The remaining 40-28.289 = 11.711 g is in excess, which is closest to the first option, that is, 11.73 grams of H₂SO₄.

6 0
3 years ago
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