357
357
Ever heard of dimensional analysis? If not, Google it. If so, What you know: 1mol potassium carbonate = 1molK2CO3 = 138.2055gK2CO3 So: 3molK2CO3 x (138.2055gK2CO3/1molK2CO3) = 414.617g potassium carbonate I'm not sure how many significant figures the 3 has (I don't know if it was 3., or 3.0 or 3.00, etc.), but if it was just 3, then your answer would be just 400g potassium carbonate.
For each molecule of KBr there is only 1 K atom (you can tell because there are no numbers in the molecule formula).
Seeing as "moles" is a measure of the number of molecules or atoms, that means each mole of KBr includes 1 mole of K.
Therefore 3.3 moles of KBr equals 3.3 moles of K (or K+ to be more accurate).
1.70 (mol KBr) * [39.10 (g/mol K+) + 79.90 (g/mol Br-)] (g/mol KBr) = 1.70*119.00 = 202.30 = 202 g KBr
Molarity = moles of solute/Liters of solution
3 M KBr = moles KBr/1 liter
= 3 moles KBr (119 grams/1 mole KBr)
= 357 grams needed
14.17 mol BaBr2 has 2*14.17 mol Br in it, so 28.34 mol KBr can be produced (also 28.34 mol K is needed)
The answer is 0,0509 mole.
There would be 0.75 moles in 1 liter of solution. You have 100 mL which is in fact 0.1 liters. so you would have 0.1 of 0.75 moles. 0.1 x 0.75 = 0.075 moles.
yes it is because kbr is just one word not 2.
The answer is 0,0207 mol.
14.17 mol BaBr2 has 2*14.17 mol Br in it, so 28.34 mol KBr can be produced (also 28.34 mol K is needed)
14.17 mol BaBr2 has 2*14.17 mol Br in it, so 28.34 mol KBr can be produced (also 28.34 mol K is needed)
0.1868 moles
0.1868 moles
Molarity = moles of solute/Liters of solution ( 25 ml = 0.025 Liters ) 1.5 M KBr = moles KBr/0.025 Liters = 0.038 moles potassium bromide ------------------
First write a balanced chemical equation: 2K + Br2 ---> 2KBR Find the limiting reactant by using the moles of each element and determining which one gives you the smallest number of moles of potassium bromide. 2.92 mol K (2 mol KBr/2 mol K)= 2.92 mol KBr 1.78 mol Br2 (2 mol KBR/1 mol Br2)=3.56 mol KBr potassium is your limiting reactant so the max. number of moles of KBr that can be produced is 2.92 mol of KBr
The answer is 0,0509 mole.
None, unless there is metallic potassium in the reaction mixture. Assuming excess potassium metal is present then 14 moles of KBr can be produced. 7BaBr2 + excess potassium -----> 14KBr + 7 Ba
0.29M = 0.29 moles in 1000 ml so 0.29 x 110/1000 = 0.0319 moles
Potassium bromide is KBr, the atomic mass of this compound is ca. 119.1. no.moles = mass/relitive molecular mass, so in this case that's 245/119.1 = 2.057 moles of KBr.
There would be 0.75 moles in 1 liter of solution. You have 100 mL which is in fact 0.1 liters. so you would have 0.1 of 0.75 moles. 0.1 x 0.75 = 0.075 moles.
Kbr