Suppose you wish to use this reaction to determine the weight percentage of TiO2 in a sample of ore. To do this you collect the O2 gas from the reaction. If you find that 1.586 grams of the TiO2 containing ore evolved 32.1 mg of oxygen gas, what is the weight percent of TiO2 and Ti in the ore

Mass percentage of TiO2 in the ore = 5.04%

Mass percentage of Ti = 3.01%

Explanation:

Equation of reaction is given below:

3 TiO2(s) + 4 BrF3(l) ---> 3 TiF4(s) + 2 Br2(l) + 3 O2(g)

From the equation of reaction above, 3 moles of O2(g) is obtained from 3 moles of TiO2(s)

Molar mass of O2 = 32 g/mol; molar mass of TiO2 = 79.87 g/mol, molar mass of Ti = 47.87 g/mol

Number of moles of O2 in 32.1 mg or 0.0321 g of O2 = 0.0321 g/32 g/mol = 0.001 moles

Therefore, 0.001 moles of O2 will be obtained from 0.001 moles of TiO2

Hence, mass l of 0.001 moles of TiO2 = 0.001 moles x 79.87g/mol = 0.07987

Mass percentage of TiO2 in the ore = 0.07987 x 100/1.586 = 5.04%

Mass of Ti in sample = (0.07987 - 0.03210) g = 0.04777 g

Mass percentage of Ti = 0.04777 × 100/1.586 = 3.01%

answer: mole of the gas

answer: mole of the gasexplanation: