What volume of 0.35 M HCl is needed to react with 5 g CaCO3? 2HCl(aq) + CaCO3(s)

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Explanation / Answer

First write out the chemical equation for the reaction: 2HCl + Ca(OH)2 -> 2H2O + 2CaCl Ca(OH)2 has a molecular mass of: Ca = 40.078 H = 1 O = 16 So we have 2 H and 2 O, which and one Ca, total molecular weight is: 40 + 2(1) + 2(16) = 74 amu 74 amu = 74 grams/moles So to find the number of moles of Ca hydroxide we have, we need to divide 25g by 74 amu which equals 0.3378378 moles. From the chemical equation at the top we can see we need twice as much hydrochloric acid as calcium hydroxide so we need 0.675675 moles of hydrogen chloride. Molarity = moles / Liters 0.900 M = moles / Liters 0.9 M = 0.675675 / Liters Liters = 0.75075 Liters of Hydrochloric acid of concentration 0.900 Molar are needed to neutralize 25g of calcium hydroxide. For the second equation we can see that the solution is saturated for when 2.47g of BaSO4 are formed when we add Na2SO4. Molecular mass of BaSO4: Ba = 137.327 S = 32 O = 16 137.3 + 32 + 16*4 = 233.3 amu So we need to figure out how many moles this is: AMU = grams / moles moles = grams/AMU moles = 2.47g/233.3 amu moles = 0.010587 moles of BaSO4 Molarity = moles / Liters Molarity = 0.010587 mol / 0.5 L Molarity = 0.02117445 M

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