How many grams of solid barium sulfate form when 35 mL of 0.160 M barium chloride reacts with 58 mL of .065 M sodium sulfate?

In a science equation, a formula is usually used to determine the mass of something. In this case, a formula is used to determine the amount of grams there are in solid barium sulfate if you combined the thirty-five milliliters of 0.160 M of barium chloride with fifty-eight milliliters of .065 M of sodium sulfate.

When these two substances react together, they form of sodium chloride in a liquid form. It would be aqueous. Sodium chloride is a fancy word for table salt. It is represented by the chemical formula NaCl. It has a molar mass of 22.99. If you had one hundred grams of sodium chloride, you would have almost 40 grams of sodium and almost 61 grams of the chloride.

.88 g-balanced chemical equaiotion:bacl2(aq)+na2so4(aq)---->baso4(s)+2nacl(aq)moles of bacl2= (35 ml)(10^-3l/1ml)(.160 mol bacl2/1l)=.00560 mol bacl2finding the moles of baso4 from the moles of bacl2:moles of baso4 from bacl2=(.00560 mol bacl2)(1 mol baso4/1 mole bacl2)=.00560 mol baso4moles of na2so4= (58 ml0 (10^-3 l/1 ml)(.065 mol na2so4/1 l)=.00377 mol baso4finding the moles of baso4=(.00377 mol na2so4)(1 mol baso4/1 mol na2so4)=.00377 mol baso4mass (g) of baso4=(.0377 mol baso4)(233.4 g baso4/1 mol baso4)=.879918=.88g baso4