BaCl2+ H2SO4 > BaSO4+ 2 HCl My assigned Vol. of BaCl2 was 15 mL And thru experim

Question

BaCl2+ H2SO4 > BaSO4+ 2 HCl My assigned Vol. of BaCl2 was 15 mL And thru experiment i found .92g of BaSO4 isolated 1. how do i calculate the theortical yield of BaSO4 using myassigned vol.? ***I dont know if these values may be any help but here theyare i used 5 mL of .6 M H2SO4 5mL of 1M of HCl BaCl2+ H2SO4 > BaSO4+ 2 HCl My assigned Vol. of BaCl2 was 15 mL And thru experiment i found .92g of BaSO4 isolated 1. how do i calculate the theortical yield of BaSO4 using myassigned vol.? ***I dont know if these values may be any help but here theyare i used 5 mL of .6 M H2SO4 5mL of 1M of HCl

Explanation / Answer

I answered the question posted after this one but I just noticedsome missing information so here it goes:
Since we do not know the limiting reactant, we must find how muchBaSO4 will be produced for both of the reactants.
Starting with H2SO4: 5 mL H2SO4 * 1 L / 1000 mL * 0.6 mol H2SO4 / 1 L H2SO4 * 1 molBaSO4 / 1 mol H2SO4 * 233.392 g/mol BaSO4 = 0.7002 g BaSO4 **NOTES** 0.6 M is the molarity of the H2SO4. 0.6 M is equal to 0.6mol/L. The molar mass of BaSO4 is 233.392 g/mol.
Next, with BaCl (I assume HCl is a typo because it is not areactant): 5 mL BaCl2 * 1 L / 1000 mL * 1 mol BaCl2 / 1 L BaCl2 * 1 mol BaSO4/ 1 mol BaCl2 * 233.392 g/mol BaSO4 = 1.167 g BaSO4    Since 0.7002 g BaSO4 < 1.167 g BaSO4, H2SO4is the limiting reactant and 0.7002 g BaSO4 is the theoreticalyield.

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