Apharmacist has calculated that a patient requires 30 mmol of phosphate and 80 meq of potassium to be added to the pn. how many milliliters of potassium phosphate and how many milliliters of potassium chloride will be required? [k-phos is 3 mmol phosphate with 4.4 meq potassium per ml and kcl is 2 meq of potassium per ml.]

Since potassium and phosphate is what we are to find for and they are both found in the potassium phosphate solution, therefore we solve for this one first on the basis of the phosphate.

The formula for finding the volume given the concentration and number of moles is:

Volume = number of moles / concentration in Molarity

Volume potassium phosphate required = 30 mmol phosphate / (3 mmol / mL)

Volume potassium phosphate required = 10 mL

This would also contain potassium in amounts of:

Amount of potassium in potassium phosphate = 10 mL (4.4 meg / mL)

Amount of potassium in potassium phosphate = 44 meg

Therefore the potassium chloride required is:

Volume of potassium chloride = (80 meg – 44 meg) / (2 meg / mL)

Volume of potassium chloride = 72 mL

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g

a 20-cm-long, cylindrical graphite (pure carbon) rod is inserted into an oxidizing atmosphere at 1145 k and 1.013 x 105 pa pressure. the oxidizing process is limited by the diffu- sion of oxygen counterflow to the carbon monoxide that is formed on the cylindrical surface.