A reaction was conducted between barium nitrate and sodium phosphate. 3 Ba(NO3)2 + 2 Na3PO4 → 6 NaNO3 + Ba3(PO4)2 What is the percent yield if 0.3 mol Ba(NO3)2 and 0.25 mol Na3PO4 react to produce 0.095 mol Ba3(PO4)2?

The % yield is 95.0 %

Explanation:

Step 1: Data given

Number of moles Ba(NO3)2 = 0.3 moles

Number of moles Na3PO4 = 0.25 moles

Number of Ba3(PO4)2 = 0.095 moles

Step 2: The balanced equation

3 Ba(NO3)2 + 2 Na3PO4 → 6 NaNO3 + Ba3(PO4)2

Step 3: Calculate the limiting reactant

For 3 moles Ba(NO3)2 we need 2 moles Na3PO4 to produce 6 moles NaNO3 and 1 mol Ba3(PO4)2

Ba(NO3)2 is the limiting reactant. It will compeltely be consumed completely (0.3 moles). Na3PO4 is in excess. There will react 0.20 moles.

There will remain 0.25-0.20= 0.05 moles

Step 4: Calculate moles Ba3(PO4)2

For 3 moles Ba(NO3)2 we need 2 moles Na3PO4 to produce 6 moles NaNO3 and 1 mol Ba3(PO4)2

For 0.3 moles Ba(NO3)2 we'll have 0.3/3 = 0.1 mol Ba3(PO4)2

Step 5: Calculate percent yield

% yield = (actual yield / theoretical yield) * 100%

% yield = (0.095 / 0.1) *100 %

% yield = 95.0 %

The % yield is 95.0 %

> b takes place in the gas phase in an isothermal tubular reactor. the feed is one mole of a per one mole of c, an inert. the entering temperature and pressure are 427°c and 10 atm, respectively. the gas constant r = 0.08206 atm · l/mol · k.