ANSWER ALL PLEASE 1.How many moles of TiCl 4 can be produced when 2.60 mol Ti re

Question

ANSWER ALL PLEASE

1.How many moles of TiCl4 can be produced when 2.60 mol Ti reacts with 8.40 mol Cl2?
mol

How many moles of each reactant are left over? (Note that at least one of the reactants must disappear

Ti remaing?

Cl2 remaing?

2. How many moles of magnesium ions, phosphate ions, and oxygen atoms are present in 0.177 mol of magnesium phosphate?

3.Find the following for the complete reaction of 3.02 mol of F2.

N2 + 3 F2 2 NF3

(a) moles of N2 required
mol

(b) mass of N2 required
g

(c) moles of NF3 produced
mol

(d) mass of NF3 produced
g

4.How many moles of O atoms are present in 0.528 moles of P4O10?


How many moles of P are in a sample of P4O10 that contains 0.484 moles of O atoms?
mol

Explanation / Answer

1.How many moles of TiCl4 can be produced when 2.60 mol Ti reacts with 8.40 mol Cl2?

mol
reaction is Ti + 2 Cl2 ...............> TiCl4

Ti : Cl2 = 1:2 to form one mole TiCl4.
there is no moles of Ti reactant left over.

moles of Cl2 left over = 8.40 - (2 * 2.6) = 3.2 mole.

Ti reactants must disappear because Ti is the limiting reactant.

2. How many moles of magnesium ions, phosphate ions, and oxygen atoms are present in 0.177 mol of magnesium phosphate?

formula of magnesium phosphate is Mg3(PO4)2.

moles of Mamgnesium ion = 3 * 0.177 = 0.531 mole

moles of phosphate ion = 2 * 0.177 = 0.354 mole.

moles of O atoms = 8 * 0.177 = 1.416 moles.

3.Find the following for the complete reaction of 3.02 mol of F2.

N2 + 3 F2 2 NF3

(a) moles of N2 required = 3.02 / 3 = 1.007 mole.
mol

(b) mass of N2 required = 1.007 * 28 = 28.19 gm
g

(c) moles of NF3 produced = 2 * 3.02 / 3 = 2.013 mole
mol

(d) mass of NF3 produced = 2.013 * 71 = 142.9 gm
g

4. moles of O atoms are present in 0.528 moles of P4O10 = 10 * 0.528 = 5.28 mole


moles of P are in a sample of P4O10 that contains 0.484 moles of O atoms = 4 * 0.484 / 10 = 0.1936 moles.

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