4 mole cuO X 2/1 = 8 mole Hcl
4 mole cuO X 2/1 = 8 mole Hcl
To find the mass of hydrogen needed to react with 40g of copper oxide (CuO), first calculate the molar mass of CuO. Then, use stoichiometry to determine the moles of CuO present in 40g. From the balanced chemical equation CuO + H2 -> Cu + H2O, you can determine the mole ratio between CuO and H2. Finally, use the molar mass of hydrogen to calculate the mass of hydrogen needed to react.
We need to go from grams of CuO to mL of H2SO4. Atomic weight of CuO = 63.55 g Cu + 16 g O = 79.55 g CuO (.80 g CuO) * (1 mol CuO / 79.55 g CuO) = .0100566 mol CuO (g CuO cancel) Since the moles of CuO is a 1:1 ratio to H2SO4 (see balanced equation) we know that: mol CuO = mol H2SO4 or 0.0100566 mol CuO = 0.0100566 mol H2SO4 3.0 M of H2SO4 means that there is 3 mol / 1 L. So we can divide this by the moles to get L then mL of H2SO4 (0.0100566 mol H2SO4) * (1 L H2SO4 / 3 mol H2SO4) * (1000 mL H2SO4 / 1 L H2SO4) = 3.4 mL H2SO4 (mol H2SO4 and L H2SO4 cancel) So 3.4 ml of H2SO4 is needed to react with 0.80 g of CuO.
To find the number of moles, first calculate the molar mass of CuO by adding the atomic masses of copper (Cu) and oxygen (O) together. Then, divide the given mass of CuO (3.2g) by the molar mass to get the number of moles.
To find the mass of 0.125 moles of copper (II) oxide (CuO), you need to determine the molar mass of CuO which is 79.55 g/mol. Multiply the molar mass by the number of moles to get the mass. Therefore, 0.125 moles of CuO would have a mass of 9.94 grams.
4 mole cuO X 2/1 = 8 mole Hcl
To find the mass of hydrogen needed to react with 40g of copper oxide (CuO), first calculate the molar mass of CuO. Then, use stoichiometry to determine the moles of CuO present in 40g. From the balanced chemical equation CuO + H2 -> Cu + H2O, you can determine the mole ratio between CuO and H2. Finally, use the molar mass of hydrogen to calculate the mass of hydrogen needed to react.
11 g hydrogen are needed.
We need to go from grams of CuO to mL of H2SO4. Atomic weight of CuO = 63.55 g Cu + 16 g O = 79.55 g CuO (.80 g CuO) * (1 mol CuO / 79.55 g CuO) = .0100566 mol CuO (g CuO cancel) Since the moles of CuO is a 1:1 ratio to H2SO4 (see balanced equation) we know that: mol CuO = mol H2SO4 or 0.0100566 mol CuO = 0.0100566 mol H2SO4 3.0 M of H2SO4 means that there is 3 mol / 1 L. So we can divide this by the moles to get L then mL of H2SO4 (0.0100566 mol H2SO4) * (1 L H2SO4 / 3 mol H2SO4) * (1000 mL H2SO4 / 1 L H2SO4) = 3.4 mL H2SO4 (mol H2SO4 and L H2SO4 cancel) So 3.4 ml of H2SO4 is needed to react with 0.80 g of CuO.
To find the number of moles, first calculate the molar mass of CuO by adding the atomic masses of copper (Cu) and oxygen (O) together. Then, divide the given mass of CuO (3.2g) by the molar mass to get the number of moles.
One mole of oxigen (O2-the diatomic molecule) is needed for 127,092 g copper to form CuO.
79.5 g of CuO = 1 g So, 3.2 g = (1*3.2) / 79.5 = 0.04 mole
To find the mass of 0.125 moles of copper (II) oxide (CuO), you need to determine the molar mass of CuO which is 79.55 g/mol. Multiply the molar mass by the number of moles to get the mass. Therefore, 0.125 moles of CuO would have a mass of 9.94 grams.
To determine the mass of CuO formed, first calculate the moles of each reactant using their molar masses. Then, determine the limiting reactant by comparing the moles of CuO that could be formed from each reactant. Finally, use the limiting reactant to calculate the mass of CuO formed based on the balanced chemical equation.
One must always first have the correct, balanced, chemical equation in order to answer general chemistry questions like this one. Once the student can answer a question such as this one, he can answer any questions since they all require the same information. The only difference in the questions is that for one question the student may need to convert grams to moles first, and for other questions the student may need to convert moles to grams first. Since the number of moles of any element or compound is directly proportional to the number of atoms or molecules, virtually any question that involves a chemical equation will ultimately require the number of moles, not grams, of each reactant and/or product. Now, let's solve this very simple question: The chemical equation must be CuCO3 ---> CO2 + CuO, and it tells us that for each mole of CuO formed, one mole of CuCO3 is consumed, therefore if we know the no. of moles of CuCO3 that reacted, then the same number of moles of CuO was formed. (I am assuming that the reaction goes to completion, that is all of the cupric carbonate reacts.) Thus, all that is needed to answer the question is to: 1) determine the number of moles of CuCO3 that reacted, and 2) calculate the number of grams of CuO that number of moles of CuCO3 would generate according to the chemical equation. The formula wt. of CuCO3 is: 63.546 + 12.011 + (3)(15.999) = 123.55 g/mol. Thus, 12.35 g of CuCO3 = 12.35 g CuCO3/123.55 g CuCO3/mol CuCO3 = 0.1000 mol CuCO3. Again, per the chemical equation, one CuO species is generated for each CuCO3 that disproportionates, therefore 0.1000 mol of CuO is created from 0.1000 mol of CuCO3. The molecular wt. of CuO is 63.546 + 15.999 = 79.545 g/mol, hence 0.1000 mol of CuO has a mass of 79.545 g/mol x 0.1000 mol = 7.954 g CuO.
The balanced equation for the reaction is: 2 Cu(s) + O2(g) -> 2 CuO(s). This shows that 2 moles of solid copper react with 1 mole of oxygen gas to produce 2 moles of solid copper II oxide.
One