A sample of rutile, an ore of titanium consisting principally of TiO2(s), was found to be 65.2% TiO2(s) by mass, with the remain
1 answer:
A sample of rutile, an ore of titanium consisting principally of TiO2(s), was found to be 65.2% TiO2(s) by mass, with the remainder being sand impurities. what is the minimum number of metric tons of the ore that must be processed to obtain 10.0 metric tons of titanium
Explanation:
Ore contains ---- 65.2%
Mass% of titanium in TiO2 can be calculated as shown below:
Given 10.0 metric tons of titanium is required.
The mass of ore that should be processed can be calculated as shown below:
Mass of Ti = ore x TiO2 % x Ti mass %
10.0 x 1000 kg = M (mass of ore) x (65.2/100) x (59.9/100) Ti
=>M=(10.0 metric tons) / (0.652 x 0.599)
=>M=25.6 metric tons
Hence, the mass of ore required is 25.6 metric tons.
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Answer:
The mass percent of aluminum sulfate in the sample is 16.18%.
Explanation:
Mass of the sample = 1.45 g
Mass of the precipitate = 0.107 g
Moles of aluminum hydroxide =
According to reaction, 2 moles of aluminum hydroxide is obtained from 1 mole of aluminum sulfate .
Then 0.001372 moles of aluminum hydroxide will be obtained from:
Mass of 0.000686 moles of aluminum sulfate :
= 0.000686 mol × 342 g/mol = 0.2346 g
The mass percent of aluminum sulfate in the sample:
7.5 M is the concentration of 60 ml of H3PO4 if it is neutralized by 225 ml of 2 M Ba(OH)2.
Explanation:
Data given:
volume of phosphoric acid, Vacid =60 ml
volume of barium hydroxide, Vbase = 225 ml
molarity of barium hydroxide, Mbase = 2M
Molarity of phosphoric acid, Macid =?
the formula for titration is used as:
Macid x Vacid = Mbase x Vbase
rearranging the equation to get Macid
Macid =
Macid =
Macid = 7.5 M
the concentration of the phosphoric acid is 7.5 M and the volume is 60 ml. Thus 7.5 M solution of phosphoric acid is used to neutralize the barium hydroxide solution of 2M.