Asked by Christy
Copper reacts with nitric acid according to the following reaction:
3 Cu (s) + 8 HNO3 (aq) --> 3 Cu(NO3)2 (aq) + 2 NO (g) + 4 H2O (l)
If an pre 1982 copper penny contains 3.10 grams of copper, what volume of 8.00 M nitric acid is required to exactly consume it? What volume of nitrogen monoxide gas measured at STP would be produced? (Remember that 1 mole of any gas measured at STP = 22.4 dm3)
3 Cu (s) + 8 HNO3 (aq) --> 3 Cu(NO3)2 (aq) + 2 NO (g) + 4 H2O (l)
If an pre 1982 copper penny contains 3.10 grams of copper, what volume of 8.00 M nitric acid is required to exactly consume it? What volume of nitrogen monoxide gas measured at STP would be produced? (Remember that 1 mole of any gas measured at STP = 22.4 dm3)
Answers
Answered by
DrBob222
Convert 3.1 g Cu into moles. moles = grams/molar mass
Using the coefficients in the balanced equation, convert moles Cu to moles HNO3.
Now convert moles HNO3, remembering M = moles/L to L.
From moles Cu, convert to moles NO formed, then moles x 22.4 L/mol = L NO at STP.
Using the coefficients in the balanced equation, convert moles Cu to moles HNO3.
Now convert moles HNO3, remembering M = moles/L to L.
From moles Cu, convert to moles NO formed, then moles x 22.4 L/mol = L NO at STP.
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