What mass of copper(II) nitrate would be produced from the complete reaction of 25.1 g of copper, according to the chemical reaction shown below?
Cu + 2 AgNO3 ----> Cu(NO3)2 + 2 Ag
i got 74.08 is that correct?
the percent of Cu in copperIInitrate is 63.5/187= about 1/3.
There for, if you react 25g Cu, you should get 75g about of copperII nitrate. Your answer looks good.
Yes, but I would round the number to 74.1 since you are allowed three significant figures from the 25.1 number.
thanks
To calculate the mass of copper(II) nitrate produced, you need to determine the molar ratio between copper and copper(II) nitrate using the balanced chemical equation.
The balanced chemical equation shows that the ratio between copper and copper(II) nitrate is 1:1. This means that for every 1 mole of copper reacted, 1 mole of copper(II) nitrate is produced.
To calculate the molar mass of copper(II) nitrate, you add up the atomic masses of all the atoms present in the compound:
Cu(NO3)2 = Cu + 2(NO3) = 1Cu + 2(14N + 3(16O)) = 63.55 g/mol + 2(14.01 g/mol + 3(16.00 g/mol)) = 63.55 g/mol + 2(14.01 g/mol + 48.00 g/mol) = 63.55 g/mol + 2(62.01 g/mol) = 63.55 g/mol + 124.02 g/mol = 187.57 g/mol.
Now we can calculate the number of moles of copper using its molar mass. The molar mass of copper is 63.55 g/mol.
Moles of copper = Mass of copper / Molar mass of copper
Moles of copper = 25.1 g / 63.55 g/mol = 0.395 moles.
Since the ratio between copper and copper(II) nitrate is 1:1, the number of moles of copper(II) nitrate produced will be the same as the moles of copper. Therefore, the mass of copper(II) nitrate produced will also be the same.
Mass of copper(II) nitrate = moles of copper(II) nitrate * molar mass of copper(II) nitrate
Mass of copper(II) nitrate = 0.395 moles * 187.57 g/mol = 73.84 g.
Therefore, the correct mass of copper(II) nitrate produced from the complete reaction of 25.1 g of copper is approximately 73.84 g, not 74.08 g.