(2) 2.16g of copper metal when treated with nitric acid followed by ig

1.	(2) 2.16g of copper metal when treated with nitric acid followed by ignition of the nitrategave 2.7 g of Cuo. In another experiment 1.15g of copper oxide upon reduction withhydrogen gave 0.92 g of copper. Show that the above data illustrates the law of definiteproportion.
Answer» Answer: Explanation: Experiment A: 2.16g of COPPER metal when treated with nitric ACID followed by ignition of the nitrate gave 2.7 g of Cuo. => For the given process, we can write the chemical REACTION as: Cu + 4HNO₃→Cu(NO₃)₂ + 2NO₂ + H₂O 2Cu(NO₃)₂ → 2CuO + 2NO₂ + O₂ => Here, one mole of copper metal will form one mole of copper oxide as the END product. atomic weight of Copper = 63g (reactant) molecular weight of copper oxide = 79g (product) ∴ CuO/Cu = 79/63 = 1.25 CuO/Cu as per the detail given for Experiment A = 2.7/2.16 = 1.25 Experiment B: 1.15g of copper oxide upon reduction with hydrogen gave 0.92 g of copper. => Chemical reaction for this process: CuO+H₂→ Cu + H₂ => In this reaction, Reactant is Copper oxide ( 63g) and product is Copper metal ( 79g) ∴ CuO/Cu = 79/63 = 1.25 According to experimental data, Cuo/Cu = 1.25 So, in both the CASE, ratio Cuo/Cu is remain same (1.25) which proves the the law of definite proportion.

(2) 2.16g of copper metal when treated with nitric acid followed by ignition of the nitrategave 2.7 g of Cuo. In another experiment 1.15g of copper oxide upon reduction withhydrogen gave 0.92 g of copper. Show that the above data illustrates the law of definiteproportion.

Answer»

Answer:

Explanation:

Experiment A: 2.16g of COPPER metal when treated with nitric ACID followed by ignition of the nitrate gave 2.7 g of Cuo.

=> For the given process, we can write the chemical REACTION as:

Cu + 4HNO₃→Cu(NO₃)₂ + 2NO₂ + H₂O

2Cu(NO₃)₂ → 2CuO + 2NO₂ + O₂

=> Here, one mole of copper metal will form one mole of copper oxide as the END product.

atomic weight of Copper = 63g (reactant)

molecular weight of copper oxide = 79g (product)

∴ CuO/Cu = 79/63 = 1.25

CuO/Cu as per the detail given for Experiment A = 2.7/2.16 = 1.25

Experiment B: 1.15g of copper oxide upon reduction with hydrogen gave 0.92 g of copper.

=> Chemical reaction for this process:

CuO+H₂→ Cu + H₂

=> In this reaction, Reactant is Copper oxide ( 63g) and product is Copper metal ( 79g)

∴ CuO/Cu = 79/63 = 1.25

According to experimental data, Cuo/Cu = 1.25

So, in both the CASE, ratio Cuo/Cu is remain same (1.25) which proves the the law of definite proportion.