Question

What mass of CU(OH)2 H2 is produced in the reaction of 7.6 grams CU(NO3)2 with 6.2 grams of KOH1CU(NO3)2+2KOH=1CU(OH)2+2KNO3

Answer 1

To find the mass of CU(OH)2 produced in the reaction, determine the limiting reactant by comparing the moles of each reactant to their stoichiometric coefficients in the balanced equation. Convert the moles of the limiting reactant to moles of CU(OH)2 using the stoichiometry, then convert the moles of CU(OH)2 to grams using its molar mass.

Step-by-step explanation:

To determine the mass of Cu(OH)2 produced in the given reaction, we need to first find the limiting reactant. To do this, we compare the number of moles of each reactant to their stoichiometric coefficients in the balanced equation. From the given masses of Cu(NO3)2 (7.6 g) and KOH (6.2 g), we can calculate the number of moles for each. Then, using the stoichiometry of the reaction, we determine how many moles of Cu(OH)2 can be produced from the limiting reactant. Finally, we convert the number of moles of Cu(OH)2 to grams using its molar mass.

Answer 2

Step-by-step explanation:

Cu(NO₃)₂ + 2 KOH ----> Cu(OH)₂ + 2 KNO₃

We have to find the mass of Cu(OH)₂ that is produced in the reaction of 7.6 g of Cu(NO₃)₂ with 6.2 g of KOH.

We have to find the limiting reagent. To do that, we will find the mass that is produceb by each reactant considering that is reacting with an excess of the other one.

Let's find the mass of Cu(OH)₂ produced by 7.6 g of Cu(NO₃)₂.

mass of Cu(NO₃)₂ = 7.6 g

molar mass of Cu(OH)₂ = 97.56 g/mol

molar mass of Cu(NO₃)₂ = 187.56 g/mol

1 mol of Cu(NO₃)₂ = 1 mol of Cu(OH)₂ ---> molar ratio

7.6 g of Cu(NO₃)₂ * (1 mol of Cu(NO₃)₂/187.56 g of Cu(NO₃)₂) * (1 mol of Cu(OH)₂/1 mol of Cu(NO₃)₂) * (97.56 g of Cu(OH)₂/1 mol of Cu(OH)₂) = 3.95 g of Cu(OH)₂

We found that 7.6 g of Cu(NO₃)₂ will produce 3.95 g of Cu(OH)₂. Let's find the mass of Cu(OH)₂ that will be produced by 6.2 g of KOH.

mass of KOH = 6.2 g

molar mass of Cu(OH)₂ = 97.56 g/mol

molar mass of KOH = 56.11 g/mol

2 mol of KOH = 1 mol of Cu(OH)₂ ---> molar ratio

6.2 g of KOH * (1 mol of KOH/56.11 g of KOH) * (1 mol of Cu(OH)₂/2 moles of KOH) * (97.56 g of Cu(OH)₂/1 mol of Cu(OH)₂) = 5.39 g of Cu(OH)₂

6.2 g of KOH will produce 5.39 g of Cu(OH)₂. So the limiting reactant is Cu(NO₃)₂ and KOH is in excess.

Answer: 3.95 g of Cu(OH)₂ is produced.