80 POINTS!! Will mark brainliest if answer is correct with all work shown. What mass in grams of tin would be required to completely react with 1.30 L of 0.750 M HBr in the following chemical reaction? Sn(s) + 4 HBr(aq) → SnBr₄ (aq) + 2 H₂ (g)

The answer is guy(Aqt)

it is correct information

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.

i believe the answer is a.