Iron burns in oxygen to form iron (III) oxide. How many moles of oxygen gas, O 2

Question

Iron burns in oxygen to form iron (III) oxide. How many moles of oxygen gas, O2, are needed to produce 2.56 moles of iron (III) oxide?

4 Fe (s) + 3 O2(g) 2Fe2O3(s)

2)

Phosphorus trichloride forms when white phosphorus reacts with chlorine. A sample of PCl3 was collected from a reaction of 77.25 g of P4 with 100g of chlorine.

P4(s) + 6 Cl2 (g) 4 PCl3(g)

A) What is the limiting reagent?

B)How much PCl3 can be obtained at the max from the given reactants?

3) How many atoms of exygen are there in 5 moles of HClO3?

4) A mixture of 10.325 grams of NH3 and 5.734 grams of oxygen gas reacts at high temperature to create nitric oxide and water.

NH3(s) +O2(s) NO(s) +H2O(s)

a) what is the limiting reagent?

b)How many grams of water can be obtained from the given reactants?

c) Calculate the mass of excess reactant left over after the reaction is completed.

Please show how to do every problem, I don't know how to do any of this. Thank you!

Explanation / Answer

1) 1 mole Fe2O3 = 3mole O2

       2.56 mole = 2.56x3 = 7.68mole

2) P4 = 124g/mol , Cl2 = 71g/mol

     a) Cl2 is the limiting agent

b) 1 mole P4 = 6 moles Cl2

77.25 g of P4 = 77.25g /124 g/mol = 0.623 mol

100g of Cl2 = 100g / 71g/mol = 1.408 mol (1.408 mol / 6 = 0.234 eqivalent to P4)

0.234 of PCl3 is possible = 0.234M x 124g/mol = 29.1 g

3) 1 mole cule HClO3 = 3 oxy atom

     1 mole = 3mole oxy atom

      5 mole = 5 x 3mole = 15 mole

4) balanced eqution is

4 NH3 + 5 O2 === 4 No + 6 H2O

NH3 = 17g/mol, so 10.325g NH3 = 0.607 mole ( need O2 of = 0.607 x 5/4 = 0.759 mole )

O2 = 32g/mol, so 5.734g/ 32g/mol = 0.179 mole ( need NH3 of = 0.179*4/5 = 0.14 mole)

a) here limiting agent is oxygen

b) 1 mole O2 = 6mole H2O

0.179 mole O2 = ? ==> 0.179 x 6 = 1.074 mole H2O = (1.074 mole x 18g/mol) = 19.33 g

c) Excess mole of NH3 left = total - yused = 0.607 - 0.14 = 0.548 mol = 0.467mol * 17 g/mol = 7.9g

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