From the calculated and the measured amounts of copper, calculate the percentage

Question

From the calculated and the measured amounts of copper, calculate the percentage error by using the following formula Error% = |Cu_cal - Cu_cop/Cu_cal| times 100% Write the ionic equation & the net ionic equation for the above reaction. The above reaction is also called Redox reaction or Oxidation - losing electrons and Reduction - gaining electrons. Which substance b oxidized and which to reduced? Calculate the theoretical yield of ammonia produced by the reaction of 100 g of H, gas and 200 g of N_2 gas. 3H_2 (g) + N_2(g) rightarrow 2NH_3(g)

Explanation / Answer

1) To Calculate the Percent Error subtract the experimental value from calculated value. Take the absolute value then divide that answer by the calculated value. Multiply the answer by 100 and add the % symbol.

2)

The ionic equation is -

2Al(s) + 3Cu 2+(aq) 2Al3+(aq) + 3Cu(s)

3)

2Al(s) + 3Cu2+(aq) -----à 3Cu(s) + 2Al3+ (aq)

Oxidation - when an element loses electrons

Al ----àAl3+ + 3e

Reduction -when an element gains electrons

Cu2+ + 2e- -----à Cu

Electrons are moving from the Al to the Cu, So Al is oxidized. Al metal end up with Al ions and Cu ions end up with Cu metal. Cu is reduced.

4)

Hydrogen reacts with nitrogen to yield ammonia according to the following equation-

3 H2(g) + N2(g) 2 NH3(g)

Convert 100 grams of hydrogen to moles of hydrogen

100 g/2.01588 g/mol = 49.606 moles

Convert 200 g of N2 to moles

200g/28.0134 g/mol = 7.139 moles

N2 is limiting reagent.

1 mole N2 produces 2 moles of NH3

7.139 moles of N2 produce 2 x 7.139 = 14.278 moles NH3

Convert moles of ammonia, to grams of ammonia:

14.278 moles x 17.03052 g/mol = 243.162 g NH3

The theoretical yield of NH3 is 243.162 grams.

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