1 mole CuO = 79.5454g CuO
3.2g CuO x 1mol CuO/79.5454g CuO = 0.040 mole CuO
112/32 moles for every mole of sulfur there are 32g
4 mole cuO X 2/1 = 8 mole Hcl
4 mole cuO X 2/1 = 8 mole Hcl
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.
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.
112/32 moles for every mole of sulfur there are 32g
4 mole cuO X 2/1 = 8 mole Hcl
4 mole cuO X 2/1 = 8 mole Hcl
79.5 g of CuO = 1 g So, 3.2 g = (1*3.2) / 79.5 = 0.04 mole
11 g hydrogen are needed.
Ar of O = 16g/mol Mr of O2 = 2(16) = 32g/mol Using the formula : mass = Mr x number of moles mass = 32g/mol x 50mols = 1600g
One mole of oxygen molecule weighs 32g. Therefore, 3 moles weigh 32 x 3= 96g
32g increased by 8% = 34.56g = 32g + (8% * 32g) = 32g + (0.08 * 32g) = 32g + 2.56g = 34.56g
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.
One mole of oxigen (O2-the diatomic molecule) is needed for 127,092 g copper to form CuO.
Grams is a unit of weight, miles is a unit of distance. There's no comparison between the two.
moles = mass/molar mass The molar mass of an oxygen atom = 16 g mol-1, as there are two oxygen atoms in diatomic oxygen this has to be doubled. 42g / 32g mol-1 = 1.3125 moles