3CU + 8HNO3 (aq) -> 3Cu(NO3)2 (aq) + 2NO (g) + 4H2O (I)if a penny contains 3.1 grams of copper, what volume of 8 M nitric acid is required to exactly consume it? What volume of nitrogen monoxide gas measured at STP would be produced?
Sorry, never covered that in Home Ed.
The balanced equation shows that 3 atoms of copper react with 8 molecules of HNO3 so 3 moles of copper also would react with 8 moles of HNO3. If we have 3.1 grams of copper, with an atomic weight of 63.54, that is 3.1/63.54, or 0.04878 mole of copper. That amount will react with 8/3 that number of moles of HNO3, so 0.04878 times 8/3 is 0.1301 mole of nitric acid. Our 8M solution contains 8 moles per liter, and we need 0.1301 mole, so we need 0.1301/8, or 0.0162 liter of 8M nitric acid solution (16.2 milliliters) to react completely with the copper. The equation shows that for every 3 atoms of copper used, we produce 2 molecules of nitric oxide (NO), so the number of moles of NO produced is 2/3 as many moles of copper. We started with 0.04878 mole of copper, so 2/3 of that is 0.03252 mole of NO. At STP one mole of a gas occupies 22.414 liters, so 0.03252 mole of NO occupies 22.414 times 0.03252, or 0.7289 liter (728.9 milliliters). Hope this answers both parts of your question.
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