2014年5月30日 · AlBr_3 + K_2SO_4 -> KBr + Al_2(SO_4)_3 How I balance equations is by first of all dividing the equation into two parts - reactants and products. Reactants Al = 3 Br = 3 K = 2 S = 1 O = 4 Products Al = 2 Br = 1 K = 1 S = 3 O = 12 Remember all atoms within the brackets get multiplied by the number outwith them. Step 1 Put a 6 before KBr AlBr_3 + K_2SO_4 -> 6KBr + Al_2(SO_4)_3 Step 2 Put a 2 ...

2015年11月27日 · The first thing I notice is that the product side has 3 sulfate groups, while the reactant side has only 1. We can multiply the potassium sulfate group by 3 so there are 3 on either side.

2016年2月5日 · 2.48mol The balanced chemical equation represents the mole ratio in which the chemicals react. Since 2 moles aluminuim are required for every 6 moles of hydrogen bromide, yet we only have 3.22 moles aluminium and 4.96 moles hydrogen bromide, it implies that HBr will get used up first and is hence the limiting reactant, while Al is in excess. So all 4.96 mol HBr is used up and reacts with 4.96 ...

2018年3月23日 · What is the equation for aluminum bromide plus chlorine gas yields aluminum chloride and bromine gas?

2016年11月21日 · First you need a stoichiometric equation: Al(s) + 3HBr rarr AlBr_3(aq) + 3/2H_2(g) Finally we get under a 1/2*L volume of dihydrogen. Each equiv of metal reduces 3 equiv of acid, and 3/2 equiv of dihydrogen gas are evolved. "Moles of aluminum" = (0.498*g)/(26.98*g*mol^-1)=0.0185*mol. Now, given the conditions, clearly the metal is the limiting reagent. We have the stoichiometry, so a molar ...