2017年4月13日 · See the explanationion below: Cl_2 + 2KBr -> 2KCl + Br_2 Given: 300g of Cl_2 300g of KBr (a) Mass of Cl_2 = 300g Molecular mass of Cl_2 = 2(35.45) = 70.906 g/mol No.of moles of Cl_2 = mass/molar mass = 300/70.906 =4.23 moles Mass of KBr = 300g Molar mass of KBr= 39+79.9 = 118.9g/mol No.of moles of KBr = 300/118.9 =2.52moles Now, Consider the equation: Cl_2 +2KBr-> 2KCl + Br_2 1 mole of Cl_2 ...

2016年1月1日 · "Cl"_2" + KBr"rarr"2KCl" + "Br"_2" There are now 2 Cl atoms on both sides. Balance the Br There are 2 Br atoms on the right side and 1 Br atom on the left side. Add a coefficient of 2 in front of KBr. "Cl"_2" + 2KBr"rarr"2KCl" + "Br"_2" The equation is now balanced with 2 Cl atoms, 2 K atoms, and 2 Br atoms on both sides of the equation.

2018年4月3日 · Cl_2 The reaction 2KBr (aq) + Cl_2 (aq) rarr 2KCl (aq) + Br_2 (l) shows that for every Cl_2 molecule needed for the reaction, 2KBr molecules are needed too. For maximal yield and no excess reactants, the molar ratio Cl_2:KBr should be 1:2.

2017年4月9日 · 188 g of Potassium Chloride This is a limiting reactant problem so we need to see which will produce the least amount of moles of potassium chloride then use that to find how many grams. 300.0 g Cl_2(1 mol Cl_2) /(70.906 g) = 4.231 mol Cl_2(2 mol KCl)/(1 mol Cl_2)=8.462 mol KCl 300.0 g KBr(1 mol KBr ) /(119.002 g) = 2.521 mol KBr(2 mol KCl)/(2 mol KBr)=2.521 mol KCl The limiting reactant is ...

2018年6月28日 · By considering that atoms must be conserved in chemical reactions, we can figure out that it must be 1Br_2 + 2KI = 2KBr + I_2 with a bit of logic. The main concept that must be applied to determine the coefficients (amount of each item) is that there must be equal amounts of each element on each side of the equation. We are not destroying or creating new atoms. In this case, the unbalanced ...

2017年11月30日 · See below. Step 1: Balance the chemical equation. Cl_2(g)+2KBr(s)->2KCl(s)+Br_2(g) Step 2: Identify the limiting reagent. This is the compound that will basically run out first, but this problem doesn't give us the amount of Cl_2 the KBr is reacting with, so we assume it is reacting in an excess of Cl_2 (meaning KBr is our limiting reagent) Step 3: Convert grams to moles using KBr molar mass ...

2018年5月15日 · "Cl"_2("g") + "2KBr(aq)"rarr"Br"_2(l) + "2KCl(aq)" This is a single replacement reaction in which a more reactive halogen (group 17/VIIA) replaces a less reactive halogen in a compound. The reactivity of the halogens relative to one another decreases down the group. Since chlorine is above bromine, it is more reactive and will replace the bromine in the potassium bromide compound. However, the ...

2017年4月6日 · First it necessary to balance all of the compounds. Then make sure that there is the same number of atoms on each side of the equation. K(s) = 0 charge. Mg = +2 charge. Br = Br-1 charge. Mg(s) = 0 charge. MgBr = Mg^+2Br_2^-1( balanced compound) KBr = K^+1Br^-1 ( balanced compound) K(s_ + MgBr_2 = KBr + Mg(s) There are two Br in the reactants but only 1 Br in the products so the KBr must be ...

Example: #"Br"_2 + "2KI" → "I"_2+ "2KBr"# To determine whether a given single replacement will occur, you must use an “Activity Series” table. If the metal or the halogen is above the element it will replace based on the activity series, a single displacement reaction will occur. Examples:

For the reaction represented by the equation #Cl_2 + 2KBr -> 2KCl + Br_2#, how many grams of potassium chloride can be produced from 300. g each of chlorine and potassium bromide? How many moles of methane do I need to react with 1 mol …