1. How many grams of sodium carbonate would be required to produce 85.0 g of barium carbonate? (the molar mass of
Na, CO, is 106 g per mole, and the molar mass of BaCO, is
197 grams per mole)

Na2CO3+ BaSO4 → BaCO3 + Na2SO4

A. 52.4 g sodium carbonate
B. 38.7 g sodium carbonate
C. 40.5 g sodium carbonate
D. 32.0 g sodium carbonate

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g

a) the particles of the chair are strongly held together.

answer carbon and three oxygen atoms;

hypostyle hall;