A student combines 1.25g of barium nitrate with an excess excess of sodium sulfate in order to make barium sulfate. What is the theoretical yield of BaSO4 for this reaction? the balance equation is

Ba(NO3)2+Na2SO4->BaSO4+2NaNO3

1.12g

Explanation:

Step 1:

The balanced equation for the reaction is given below:

Ba(NO3)2 + Na2SO4 —> BaSO4 + 2NaNO3

Step 2:

Let us calculate the mass of barium nitrate, Ba(NO3)2, that reacted and the mass of barium sulfate, BaSO4, produced from the balanced equation. This is illustrated below:

Molar Mass of of barium nitrate, Ba(NO3)2 = 137 + 2[14 + (16x3)] = 137 + 2[14 +48] = 137 + 2[62] = 137 + 124 = 261g/mol

Molar Mass of barium sulfate, BaSO4 = 137 + 32 + (16x4) = 137 + 32 + 64 = 233g/mol

Step 3:

Determination of the theoretical yield of barium sulfate, BaSO4. The theoretical yield of barium sulfate, BaSO4 can be obtained as follow:

From the balanced equation above, 261g of Ba(NO3)2 produced 233g of BaSO4.

Therefore, 1.25g of Ba(NO3)2 will produce = (1.25 x 233)/261 = 1.12g of BaSO4.

Therefore, the theoretical yield of barium sulfate, BaSO4 is 1.12g

answer; i believe the correct answer is (solid gold);

i believe you are right for all except the 3rd ( look at sqdancefan his answer is much better)