1g mixture of Cu2O and CuO on heating gives 0.88g of copper. Calculate % of CuO in mixture
Well, let's break it down. On heating, the copper in Cu2O and CuO will react and convert to copper metal (Cu).
So, if we assume that all the copper in the mixture comes from CuO only, then the 0.88g of copper is from CuO.
To find the percent of CuO in the mixture, we use the formula:
% CuO = (mass of CuO / total mass of mixture) * 100
First, let's calculate the mass of CuO in the mixture:
mass of CuO = 0.88g
Now, let's calculate the total mass of the mixture:
total mass of mixture = 1g
Now, we can substitute these values into the formula:
% CuO = (0.88g / 1g) * 100
% CuO = 88%
So, the percentage of CuO in the mixture is 88%. Let's hope the copper didn't get too hot and become too volatile. We wouldn't want a metallic tantrum now, would we?
To calculate the percentage of CuO in the mixture, we need to first determine the mass of CuO that is present in the mixture.
Given:
Mass of the mixture = 1g
Mass of copper obtained = 0.88g
To find the mass of CuO:
Let x be the mass of CuO.
Therefore, the mass of Cu2O can be calculated as (1 - x) g.
Now, we need to calculate the moles of CuO and Cu2O using their respective molar masses:
Molar mass of CuO = 63.55 + 16 = 79.55 g/mol
Molar mass of Cu2O = (2 * 63.55) + 16 = 143.1 g/mol
Next, we'll calculate the moles of CuO and Cu2O present in the mixture:
Moles of CuO = (x / 79.55)
Moles of Cu2O = ((1 - x) / 143.1)
Since CuO and Cu2O react to form copper, the moles of CuO and Cu2O should be in a 1:1 ratio. Therefore, we can set up the following equation:
Moles of CuO / Moles of Cu2O = (x / 79.55) / ((1 - x) / 143.1) = 1
Simplifying the equation gives:
(x / 79.55) = ((1 - x) / 143.1)
Now, we'll solve this equation to find the value of x:
143.1x = 79.55 - 79.55x
143.1x + 79.55x = 79.55
222.65x = 79.55
x = 79.55 / 222.65
x = 0.357 g
Hence, the mass of CuO in the mixture is 0.357 g.
Finally, we can calculate the percentage of CuO in the mixture:
% of CuO = (mass of CuO / mass of mixture) * 100
% of CuO = (0.357 / 1) * 100
% of CuO = 35.7%
Therefore, the percentage of CuO in the mixture is 35.7%.
To calculate the percentage of CuO in the mixture, we first need to determine the amount of CuO and Cu2O that reacted to form copper.
Let's assume the mass of CuO in the mixture is x grams.
So, the mass of Cu2O in the mixture would be (1 - x) grams.
According to the given information, when the mixture is heated, it gives 0.88g of copper. Since the molar mass of copper is the same regardless of the oxidation state, we can use stoichiometry to calculate the amount of CuO and Cu2O that reacted.
The balanced equation for the reaction is:
2CuO + Cu2O -> 6Cu + O2
From the equation, we can see that 1 mole of CuO reacts with 1/2 mole of Cu2O to give 3 moles of copper. Therefore, the number of moles of CuO can be calculated as follows:
moles of CuO = (0.88g of Cu) / (molar mass of Cu)
Next, we need to find the number of moles of Cu2O that reacted. Since the molar mass of Cu2O is twice that of CuO, we can use the previously calculated moles of CuO to find the moles of Cu2O.
moles of Cu2O = 0.5 * moles of CuO
Now, we can calculate the mass of CuO and Cu2O using their respective moles:
mass of CuO = moles of CuO * (molar mass of CuO)
mass of Cu2O = moles of Cu2O * (molar mass of Cu2O)
Finally, we can calculate the percentage of CuO in the mixture:
% CuO = (mass of CuO / total mass of mixture) * 100
By substituting the values obtained from the calculations, we can find the percentage of CuO in the mixture.
Two equations in two unknowns and you solve them simultaneously.
2CuO ==> 2Cu + O2
2Cu2O ==> 4Cu + O2
-----------------
Let x =g CuO
and y = g Cu2O
----------------
eqn 1 is x + y = 1g
eqn 2. This converts Cu from CuO and Cu from Cu2O to grams Cu that you end up with of 0.88 g. mm stands fr molar mass. am stands for atomic mass.
x(2*am Cu/2*mm CuO) + y(4*am Cu/2*mm) = 0.88 Cu2O
Solve the two equations simultaneously for x and y. x gives you grams CuO.
Then %CuO = (g CuO/1 g)*100 = ?
Post your work if you have trouble.