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%

what's the

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2