A reaction produced 52.1 g of kcl how much o2 was produced in grams?
Equation: 2KClO3 → 2KCl + 3O2 If you produce 2 mol KCl you produce 3 mol O2 Molar mass KCl = 39.1+35.5 = 74.6g/mol Mol KCl in 52.1g = 52.1/74.6 = 0.698 mol KCl From the equation , you will produce 0.698*3/2 = 1.048 mol O2 Molar mass O2 = 32g/mol Mass O2 produced = 32*1.048 = 33.5g O2 produced.
If this reaction produced 77.2g of KCl, how much O2 was produced (in grams)?
1.(77.2g KCl)/(74.55 grams/ mole KCl)= 1.036 moles of KCl 2.(1.036 moles of KCl)x (3 moles of O2)/(2 moles of KCl) = 1.55 moles of O2 3.(1.55 moles of O2)(32.00 grams/ mol O2) = 49.7 grams of O2
If this reaction produced 89.2 g of KCL, how much O2 was produced (in grams)?
Mass of KCl=89.2g Molar mass of KCl=39+35.5 =74.5 Moles of KCl= 89.2/74.5 Moles of KCl=1.197 moles For 2 moles of KCl there are 3 moles of oxygen 1 mole of KCl give=3/2=1.5 moles of oxygen 1.197 moles of KCl give =1.5*1.197 Moles of oxygen=1.796 moles Molar mass of O2=32 Mass=1.796*32 Mass of oxygen=57.47 g Hope it helps:) In one of your questions i deleted my answer and came to know that i can not answer it again.but its answer is as follows First of all we will find the moles of octane Mass of octane=18.6g Molar mass of octane =12*8+18 = 114 Moles of octane=mass /molar mass = 18.6/114 Moles of octane=0.1632 moles 2 moles of octane give 16 moles of CO2 as given in equation 1 mole of octane gives 16/2=8 moles of CO2 0.1632 moles of octane give=8*0.1632 Moles of CO2=1.31 moles Hope it helps :) This is the right answer!!
If this reaction produced 20.1 g of KCl, how much O2 was produced (in grams)?
2 KClO3 => 2 KCl + 3 O2, and we know we get 20.1g of KCl. First, we have to convert 20.1g of KCl into moles of KCl. The molar mass of KCl is 74.55g/mol. So, 20.1g KCl / 74.55g KCl/mol = 0.270 mol KCl. Next, we have to convert moles of KCl to moles of O2. We know from the equation that 2 moles of KCl gives 3 moles of O2. So, 0.270 mol KCl * 2 mol KCl / 3 mol O2 = 0.180 mol O2. Finally, we have to convert moles of O2 into grams of O2. Since the molar mass of O2 is 32g/mol, 0.180g O2 * 32g/mol = 5.76g O2. And there ya go!
If this reaction produced 74.9g of kcl, how much o2 was produced (in grams)?
Convert to moles: 74.9g KCl / 74.6 g/mol KCL = 1.00 mols KCL The molar ratio KCL : O2 = 2 : 3 = 1 : x; x = 1.5 moles O2 1.5 moles O2 x 32 g/mol = 48 g O2 Cool!
How many grams of KClO3 were decomposed to produce 14.3 grams of O2?
I don't have calculator nearby, but you can look for term, oxygen balance. Its a simple formula which will tell you free oxygen liberated by KP which can act as oxidizer. Ammonium perchlorate has oxygen balance of +34%.
How many grams of Cl2 can be prepared from the reaction of 30.0 g of HCl with excess MnO2 according to the following chemical equation? MnO2+4HCl→MnCl2+Cl2+2H2O why is Answer: 13.0g?
The answer should be 14.6g.30g HCl = .823mol.823mol HCl => .206mol Cl2 (Cl2 is produced in a 1:4 ratio).206mol Cl2 = 14.6gIf the question has "percent yield" in it somewhere, then if your ideal yield is only 89% your maximum yield will be 13g.To check your answer, here's a stoichiometry calculator I made back when I was a sophomore.Reaction Stoichiometry CalculatorNote: You can never go wrong with tried and true grams to moles, moles to grams, but Debarshi's method is a little easier.
How many grams of KCl are produced from the decomposition of 56.7 grams of KClO3?
First, write the balanced equation:[math]2KClO_3 \xrightarrow{\text{heat}} 2 KCl + 3O_2 [/math]Get the [math]M_r[/math] of [math]KClO_3[/math] and [math]KCl[/math]([math]122.55\text{ g mol}^{-1}[/math] and [math]74.6\text{ g mol}^{-1}[/math] respectively) Next, find out how many moles of [math]KClO_3[/math] are present. From calculations ([math]n=m/Ar[/math]), there are 0.463 mol of [math]KClO_3[/math] present.From the above, we can infer that 2 mol of [math]KClO_3[/math] decompose to give 2 mol of [math]KCl[/math]. So, by the complete thermal decomposition of 0.463 mol of [math]KClO_3[/math], we get 0.463 mol of [math]KCl[/math].0.463 mol of [math]KCl[/math] is equal to 34.5 g of [math]KCl[/math]. Hence,34.5 g of [math]KCl[/math] was produced.
Finding grams in reactions (Please help me, I don't know how to figure out the answers!)?
The coefficients are the amount of moles of each substance made for each of the other amount of moles. Ex: For every 2 moles of KCl, 3 moles of O2 are made and 2 moles of KClO3 are used. By keeping that in mind, you can set up ratios to calculate the actual amount of moles produced and easily convert it back to grams. So 14.6 g KCl / MW = moles KCl mole KCl * (3 mole O2 / 2 mole KCl) = moles O2. Convert that back to grams and you have your answer. The same process goes for the other one.
How many grams of AlCl3 will be produced if 2.50 moles of Al react?
This is a very simple chemistry question, which sounds alot like a homework question I did a while ago ;)Assuming this is a homework question, it doesn't seem that you need to deal with things like impurities, the reactants aren't any special isotopes, you wont be using significant digits, that aluminum is the limiting reactant ect. anyways, here is how its done.Your general equation is Al + 3Cl -> AlCl3A mole of Aluminum weighs 26.98 grams. 26.98 x 2.5 = 67.45 (so that is the weight of the aluminum in your AlCl3 compound. Now lets add the weight of the Cl.the molar mass of chlorine is 35.45 grams/mole. This is where it can get tricky. According to your general equation, you will be using 3 moles of Cl for every mole of Al. So you will actually be calculating the weight of 7.5 moles of Cl.35.45 x 7.5 = 265.875. Lets add up the total weight of your product now.256.875 + 67.45 = 333.415 grams.Therefore you will have 333.415 grams of AlCl3 if you react 2.5 moles of aluminum completely with Cl2 (a diatomic gas) to create AlCl3