how many grams of nah2po4 are needed to react with 36.29 ml of 0.250 m naoh?
nah2po4(s) + 2 naoh(aq) → na3po4(aq) + 2 h20(1)

The answer to your question is: 639.6 g of NaH2PO4

Explanation:

Data

NaH2PO4   = ?

NaOH 36.29 ml of  0.250 M

MW Na2PO4 = (23 x 2) + (1 x 31) + (16 x 4) = 141 g

MW NaOH = (23 x 1) + (16 x 1) + (1 x 1) = 40 g

Molarity = # moles / volume

# moles = molarity x volume

# moles of NaOH = 0.25 x 36.29 = 9.07

                    1 mol of NaOH  40 g

                  9.07 moles           x

        x = 9.07 x 40 / 1 = 362.9 g of NaOH

             NaH2PO4(s) + 2 NaOH(aq)   →   Na3PO4(aq) +   2 H20(1)

        141 g of NaH2PO4     2(40) g of NaOH

             x                             362.9 g of NaOH

                  x = 362.9 (141) / 80

                 x = 639.6 g of NaH2PO4

the reaction mixture contains some ammonia, plus a lot of unreacted nitrogen and hydrogen. the mixture is cooled and compressed, causing the ammonia gas to condense into a liquid. the liquefied ammonia is separated and removed. the unreacted nitrogen and hydrogen are then recycled back into the reactor .

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3