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Title: AP CHEMISTRY HELP! An 8.00 g sample of a copper ore was dissolved in a acid solution and then reacted with? Post by: oblivious122 on Aug 21, 2012 NO3- from KNO3 salt solution.
3Cu +8NO3- + 8H+-------->3Cu+2 + @NO +4H2O + 6NO3- It required 71.5 ml of a 0.155M KNO3 solution to fully react the copper in the ore sample. What is a percent copper in the ore? Title: AP CHEMISTRY HELP! An 8.00 g sample of a copper ore was dissolved in a acid solution and then reacted with? Post by: Tom291 on Aug 21, 2012 Step One
======= Find the moles of NO3- on the left side of the equation. v = 71.5 mL = 0.0715 L C = 0.155 M n = C*V = 0.0111 moles Step Two ======= Find the number of moles of copper 3 moles of copper requires 8 moles of N03- x moles of copper requires 0.0111 moles of N03- 8x = 3*0.0111 8x = 0.0333 x = 0.00416 moles of copper Step Three ======== Find the grams of copper. n = 0.00416 mol given mass = x Molar Mass = 63.5 grams / mol x = n*MM x = 0.00416 mol * 63.5 = 0.2639 grams Step Four ======= Find the % in the sample % = (given mass/Sample Mass)*100 % = (0.2639/8.00) * 100 = 3.30% Title: AP CHEMISTRY HELP! An 8.00 g sample of a copper ore was dissolved in a acid solution and then reacted with? Post by: fisher 2012 on Aug 21, 2012 First calculate the moles of NO3- (KNO3) from volume and molarity of KNO3.
Moles KNO3 = molarity x L of KNO3 = (0.155 M) (0.0715 L) = (0.155 mole KNO3/L) (0.0715 L) = 0.0111 mole KNO3 Which means the solution contains 0.0111 mole K+ and 0.0111 mole NO3- ions. Use the stoichiometry from the equation to calculate the moles Cu and grams Cu using the following setup. 0.0111 mole NO3- (3 mole Cu/8 mole NO3-) (63.55 g Cu/1 mole Cu) = 0.265 g Cu This means that 8.00g of ore contains 0.265 g Cu. Precent Cu in the ore = (0.265 g /8.00 g) x 100% = 3.31% |