Note that 2 mol dmā»Ā³ = 2 mol/(dmĀ³)outline: we're given the concentration of hcl, which is 2 mol dmā»Ā³ (cubic meters in the denominator can represent volume, so i know this is concentration). we're an amount of calcium carbonate cacoā, which is 5 g. we're presented with a chemical equation which includes cacoā and hcl. we are told to find the volume of hcl. from what we see so far, this seems to be a stoichiometry problem. we'll have to use the given information about calcium carbonate to find the amount of moles of hcl needed to fully react with calcium carbonate. then we'll have information about the moles of hcl and the concentration of hcl; that will be enough to figure out volume according to the formula concentration = moles / volume.======all the answer choices are not in dm.convert the given concentration so that it uses cmā»Ā³.we can use conversion factors. we have to cube our conversion factors so that dimensional analysis works.we can convert to meters first, using the fact that 10 dm = 1 m ā 10Ā³ dmĀ³ = 1 mĀ³then we use the fact that 10Ā² cm = 1 m ā (10Ā² cm)Ā³ = (1 m)Ā³ ā 10ā¶ cmĀ³ = 1 mĀ³so converting:
now we have to find the moles of hydrochloric acid neededfind the moles of calcium carbonate. we have to find the molar mass of calcium carbonate, cacoā, which can be found by molar mass = (mass ca) + (mass c) + 3(mass o) = 40 + 12 + 3(16) = 100 g/molconvert the given 5 g to moles: mol cacoā = 5 g cacoā * 1 mol/(100 g) = 0.05 mol cacoāuse stoichiometry (ratios) to figure out the moles of hcl needed. ratio wise, there's 2 hcls for every cacoā, so use dimensional analysis mol hcl = mol cacoā * (2 mol hcl) / (1 mol cacoā) = 0.05 mol hcl * 2 = 0.1 mol hcl we have 0.1 mol hcl and concentration of hcl (0.002 mol/cmĀ³).we need to find volumesince concentration = moles / volume, rearrange to get volume = moles / concentration volume = 0.1 mol / (0.002 mol/cmĀ³) volume = 50 cmĀ³
