Given the reaction: 3 Cu + 8 HNO3 --> 3 Cu(NO3)2 + 2 NO + 4 H2O
If 12.0 moles of copper react, how many grams of HNO3 are needed?
what is 12*8/3 moles?
what is the above moles multiplied by the formula mass of Nitric acid?
To determine how many grams of HNO3 are needed to react with 12.0 moles of copper, we need to use stoichiometry. Stoichiometry is a way to relate the ratios of reactants and products in a chemical reaction.
First, let's find the molar ratio between Cu and HNO3 from the balanced chemical equation:
3 Cu + 8 HNO3 → 3 Cu(NO3)2 + 2 NO + 4 H2O
From the equation, we can see that for every 8 moles of HNO3, we need 3 moles of Cu.
Now we can set up a proportion to find the amount of HNO3 needed:
(8 moles of HNO3 / 3 moles of Cu) = (x moles of HNO3 / 12.0 moles of Cu)
Cross-multiplying, we get:
8 moles of HNO3 * 12.0 moles of Cu = 3 moles of Cu * x moles of HNO3
96 moles of HNO3 = 3x moles of HNO3
Dividing both sides by 3, we find:
32 moles of HNO3 = x moles of HNO3
Therefore, 32 moles of HNO3 are needed to react with 12.0 moles of copper.
To convert moles to grams, we need to use the molar mass of HNO3. The molar mass of HNO3 is:
1 atom of H * 1 g/mol + 1 atom of N * 14 g/mol + 3 atoms of O * 16 g/mol = 63 g/mol.
So, to find the mass of HNO3, we multiply the number of moles (32 moles) by the molar mass:
Mass of HNO3 = 32 moles of HNO3 * 63 g/mol = 2016 grams of HNO3.
Therefore, 2016 grams of HNO3 are needed to react with 12.0 moles of copper.