aaronstalnaker
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Suppose that a solution containing 3.50 grams of sodium phosphate (Na3PO4) is mixed with a solution containing 6.40 grams of barium nitrate (Ba(NO3)2). Calculate the theoretical yield of barium phosphate (Ba3(PO4)2) during this reaction.
2Na3PO4 + 3Ba(NO3)2 Ba3(PO4)2 + 6NaNO3

Respuesta :

(3.50 g Na3PO4) / (163.9408 g Na3PO4/mol) = 0.021349 mol Na3PO4

(6.40 g Ba(NO3)2 / (2261.3377 g Ba(NO3)2/ mol = 0.024489 mol Ba(NO3)2

(0.024489 mol Ba(NO3)2 x 1/3)  x (601.9261 g Ba3(PO4)2/mol) = 4.91 g Ba3(PO4)2

PLATO USERS answer is B

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