]how many grams of ba are needed to react completely with 1.30 l of a 6.00 m hbr solution? ba + 2hbr babr2 + h2 3.90 g 7.80 g 534 g 632 g

536 grams  
The balanced equation is Ba + 2 HBr ==> BaBr2 + H2, so that indicates that mole of Ba will react with 2 moles of HBr. So let's determine how many moles of HBr we have and from there, determine the number of moles of Ba needed and the mass of that much Ba. The number of moles of HBr is simply the product of volume of solution used and its molarity. So: 
1.30 L * 6.00 mol/L = 7.80 mol  
And since we only need half as many moles of Ba, 7.80 mol / 2 = 3.90 mol of Ba is needed. The atomic weight of Ba is 137.327 g/mol. So the mass needed is 137.327 g/mol * 3.90 mol = 535.5753 g.  
Rounding to 3 significant figures gives 536 grams.

The answer is c. 320 g