mass of KClO3 = 66.8 g
molar mass of KClO3 = 122.55 g/mol
mol of KClO3 = (mass)/(molar mass)
= 66.8/122.55
= 0.5451 mol
According to balanced equation
mol of O2 formed = (3/2)* moles of KClO3
= (3/2)*0.5451
= 0.8176 mol
mass of O2 = number of mol * molar mass
= 0.8176*32
= 26.2 g
Answer: 26.2 g
I am allowed to answer only 1 question at a time
Small quantitles of axygen can be prepared in the laboratory by heating potassum chlorate, KCios(9 The...
Small quantities of oxygen can be prepared in the laboratory by heating potassium chlorate, KCIO, (s). The equation for the reaction is 2 KCI +302 2 KCIO Calculate how many grams of 0, (g) can be produced from heating 11.6 g KCIO, (s). mass:
Small quantities of oxygen can be prepared in the laboratory by heating potassium chlorate, KClO3(s). The equation for the reaction is 2KClO3⟶2KCl+3O2 Calculate how many grams of O2(g) can be produced from heating 98.6 g KClO3(s). mass:
Small quantities of oxygen can be prepared in the laboratory by heating potassium chlorate, KClO3(s). The equation for the reaction is 2KClO3⟶2KCl+3O2⟶2KCl+3O2 Calculate how many grams of O2(g) can be produced from heating 48.8 g KClO3(s).
Small quantities of oxygen can be prepared in the laboratory by heating potassium chlorate, KClO3(s) . The equation for the reaction is 2KClO3⟶2KCl+3O2 Calculate how many grams of O2(g) can be produced from heating 70.4 g KClO3(s) STRATEGY: Convert the mass of KClO3 to moles. Convert the number of moles of KClO3 to the number of moles of O2 . Convert the number of moles of O2 to grams. Step 1: 70.4 g KClO3 is equal to 0.574 mol KClO3...
Small quantities of oxygen can be prepared in the laboratory by heating potassium chlorate, KCIO,(s). The equation for the reaction is 2 KCIO, 2 KCI+ 30, Calculate how many grams of 0, (g) can be produced from heating 68.1 g KCIO,($). mass Iodine is prepared both in the laboratory and commercially by adding Cl, (g) to an aqueous solution containing 2 Nal(aq) + Cl2(8) — 1(s) + 2 NaCl(aq) How many grams of sodium iodide, Nal, must be used to...
1. Small amounts of oxygen gas can be produced for laboratory use by heating potassium chlorate, which causes it to decompose by the following reaction: ___KClO3(s) → ____KCl(s) + ___O2(g) (unbalanced) Balance the equation, and determine the mass of oxygen that will be formed if 15.0 g of potassium chlorate decomposes. a. 11.7 g b. 57.5 g c. 173 g d. 5.88 g e. 86.1 g
Oxygen gas can be prepared in the laboratory by heating potassium chlorate, so that it decomposes according to the equation: What would the pressure of O2 be in a 2.50 L glass container at 220°C if 6.50 g of KCIO3 is decomposed? Multiple Choice 0.0796 atm 178 atm 662 atm < Prev 6 of 16 Nht >
Oxygen, O_2 can be prepared by heating potassium chlorate, KCIO_3. 2 KCIO_3 rightarrow 2 KCI +3 O_2 How many grams of O_2 will be produced when 22.1 g of KCIO_3 are heated? If 6.79 g of oxygen are recovered, what is the percent yield?
Question 7 of 24 > Gaseous hydrogen and oxygen can be prepared in the laboratory from the decomposition of gaseous water. The equation for the reaction is 2H,0(g) — 2H,(8) + O2(g) Calculate how many grams of O2(g) can be produced from 61.9 g H,O(g). mass: go,
For the reaction 2 KI + Pb(NO), — Pol, + 2 KNO, how many grams of lead(II) nitrate, Pb(NO3)2, are needed to react completely with 39.3 g of potassium iodide, KI? mass: 2 KI + Pb(NO3)2 Pol, + 2 KNO how many grams of lead(II) nitrate, Pb(NO3)2, are needed to react completely with 39.3 g of potassium iodide, KI? mass: Attem Small quantities of oxygen can be prepared in the laboratory by heating potassium chlorate, KCIO,(s). The equation for the...