How many moles of baso4 are formed if 0.5 moles of na2 so4 react with 60 g of bacl2? na2 so4 + bacl2 → baso4 + 2nacl

the balanced equation for the reaction is as follows
Na₂SO₄ + BaCl₂ > 2NaCl + BaSO₄ 
stoichiometry of Na₂SO₄ to BaCl₂ is 1:1
first we need to find out which the limiting reactant is
limiting reactant is fully used up in the reaction.
                                                                           
number of Na2So4 moles - 0.5 mol number of BaCl2 moles - 60 g / 208 g/mol = 0.288 mol
since molar ratio is 1:1 equal number of moles of both reactants should react with each other
                                                                         
therefore BaCl2 is the limiting reactant and Na2SO4 is in excess. amount of product formed depends on number of limiting reactant present.
stoichiometry of BaCl₂ to BaSO₄ is 1:1.
                                                                           
therefore number of BaSO4 moles formed - 0.288 mol

0.29

Explanation:

To calculate the moles, we use the equation:

Putting values in above equation, we get:

Moles of = 0.5 moles

According to stochiometry, 1 mole of reacts with 1 mole of

Thus 0.29 moles of will react with 0.29 moles of

Thus is the limiting reagent as it limits the formation of products and is the excess reagent.

1 mole of produces = 1 mole of

0.29 moles of will produce= 0.29 moles of

Both the dipole-dipole and hydrogen are essentially same because they result form formation of partial negative and positive charges on the atoms constituting a given molecule. in such molecule, one atom is more electronegative than the other and therefore pulls the shared electrons more to itself than the other atom. the atom that pulls a pair of electrons more to itself becomes partially negatively charged whereas the other atom becomes partially positively charge. the weak negative charges on one molecule attracts weak positive charge on the other molecule creating intermolecular forces referred to as dipole-dipole bonds. Hydrogen bonds form in the same manner they are stronger than typical dipole-dipole bonds. so according to the explanation above  D is the correct answer.