How many grams of solid BaSO4 will form when Na2SO4 reacts with 25 mL of 0.50 M Ba(NO3)2? __ Ba(NO3)2 + __ Na2SO4  __ BaSO4 + __ NaNO3

Question

Caleb Adams · Accepted Answer

The answer is #2.79g#. Initially focusing on the chemical equation in balance #Ba(NO_3)_(2(aq)) + Na_2SO_(4(aq)) -> BaSO_(4(s)) + 2NaNO_(3(aq))# Taking into consideration the solubility rules (more here: https://tutor.hix.ai), we can see that #Ba(NO_3)_2#, #Na_2SO_4#, and #NaNO_3# will dissociate into their respective ions, which will lead to the reaction's net ionic equation #Ba_((aq))^(2+) + SO_(4(aq))^(2-) -> BaSO_(4(s))# Since sulfate compounds formed with #Ba^(2+)# cations are insoluble in water (only slightly soluble, #BaSO_4#'s #K_(sp)# being equal to #1.1 * 10^(-10)# ), this double-replacement reaction forms a precipitate, #BaSO_4#. We know from the balanced chemical equation that the mole-to-mole ratio of #Ba(NO_3)_2# and #BaSO_4# is 1:1; that is, for every mole of barium nitrate used, one mole of barium sulfate is produced.  The number of barium nitrate moles can be determined from its molarity, #C = n/V# #n_(Ba(NO_3)_2) = C * V = 0.500 M * 25.0 * 10^(-3) L = 0.012#  Knowing barium sulfate's molar mass (#233.3 g/(mol)#), and the number of moles produced, we get #m_(BaSO_4) = n_(BaSO_4) * molarmass = 0.012 mol es * 233.3g/(mol e) = 2.79g#

Caleb Adams · Answer

undefined The balanced chemical equation for the reaction is:

Ba(NO3)2 + Na2SO4 → BaSO4 + 2 NaNO3

From the equation, we see that 1 mole of Ba(NO3)2 reacts with 1 mole of Na2SO4 to produce 1 mole of BaSO4.

To find the number of moles of Ba(NO3)2 reacting, we use the formula:

moles = concentration × volume (in liters)

Given that the volume of Ba(NO3)2 is 25 mL (which is 0.025 L) and the concentration is 0.50 M, we can calculate the number of moles of Ba(NO3)2:

moles of Ba(NO3)2 = 0.50 M × 0.025 L = 0.0125 moles

Since the stoichiometric coefficient of Ba(NO3)2 is 1 in the balanced equation, it reacts with 1 mole of Na2SO4.

Therefore, the number of moles of BaSO4 formed is also 0.0125 moles.

Now, to find the mass of BaSO4 formed, we use its molar mass, which is approximately 233.39 g/mol.

mass of BaSO4 = moles × molar mass
                = 0.0125 moles × 233.39 g/mol
                ≈ 2.92 grams

So, approximately 2.92 grams of solid BaSO4 will form.

How many grams of solid BaSO4 will form when Na2SO4 reacts with 25 mL of 0.50 M Ba(NO3)2? __ Ba(NO3)2 + __ Na2SO4 __ BaSO4 + __ NaNO3

How many grams of solid BaSO4 will form when Na2SO4 reacts with 25 mL of 0.50 M Ba(NO3)2? Ba(NO3)2 + Na2SO4 BaSO4 + NaNO3