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A-R-Q
 Post subject: [SCH] Mixture problem (1)
PostPosted: Sat, 16 Jul 2011 17:18:42 UTC 
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A sample of mixture containing only CuO & Cu2O has mass of 1.512g. When this sample reacted with excess hydrogen gas (H2), 1.275g of copper metal (CU(s)) formed.

What is the mass percent of CuO & Cu2O in the original mixture?

I know the equation can be set up as this:

CuO + Cu2O + H2 -----> Cu(s) ......

MM CuO = 143.10g CuO / mol CuO
MM Cu2O = 79.55g Cu2O / mol Cu2O
MM Cu(s) = 63.55g Cu / mol Cu

I let "x" represent the mol of CuO
I let "y" represent the mol of Cu2O

The problem I am having is cancelling out the units in the first equation. This is how I thought it would be:

[1]

x mol CuO + y mol Cu2O = 1.512 g (CuO + Cu2O)
x mol CuO (75.99g CuO / mol CuO) + y mol Cu2O (143.10g Cu2O / mol Cu2O) = 1.512g (CuO + Cu2O)

75.99x g CuO + 143.10y gCu2O = 1.512 g(CuO + Cu2O)

I wasn't able to cancel out the units here. How would you do it?


But for the second equation,

[2] I know that all of the hydrogen gas reacted with the two copper compounds and still remained, so that means all the copper transferred from the two compounds to that one metallic solid form.


molCu (s) = 1.275g Cu(s)*(1 mol Cu(s) / 63.55g Cu(s) )
molCu (s) = 0.02mol Cu(s)


x mol Cu + 2y mol Cu = 0.02 mol Cu
Here I was able to cancel out the units:

mol Cu (x + 2y) = (0.02) mol Cu
x + 2y = 0.02

But the real problem is the first equation. How would I go about with canceling the units in the first equation?

because:

x mol CuO * (1 mol Cu /1 mol CuO)

y mol Cu2O * (2 mol Cu / 1 mol Cu2O)

[/quote]
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outermeasure
 Post subject: Re: [SCH] Mixture problem (1)
PostPosted: Sat, 16 Jul 2011 17:26:28 UTC 
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A-R-Q wrote:
A sample of mixture containing only CuO & Cu2O has mass of 1.512g. When this sample reacted with excess hydrogen gas (H2), 1.275g of copper metal (CU(s)) formed.

What is the mass percent of CuO & Cu2O in the original mixture?

I know the equation can be set up as this:

CuO + Cu2O + H2 -----> Cu(s) ......

MM CuO = 143.10g CuO / mol CuO
MM Cu2O = 79.55g Cu2O / mol Cu2O
MM Cu(s) = 63.55g Cu / mol Cu

I let "x" represent the mol of CuO
I let "y" represent the mol of Cu2O

The problem I am having is cancelling out the units in the first equation. This is how I thought it would be:

[1]

x mol CuO + y mol Cu2O = 1.512 g (CuO + Cu2O)
x mol CuO (75.99g CuO / mol CuO) + y mol Cu2O (143.10g Cu2O / mol Cu2O) = 1.512g (CuO + Cu2O)

75.99x g CuO + 143.10y gCu2O = 1.512 g(CuO + Cu2O)

I wasn't able to cancel out the units here. How would you do it?


But for the second equation,

[2] I know that all of the hydrogen gas reacted with the two copper compounds and still remained, so that means all the copper transferred from the two compounds to that one metallic solid form.


molCu (s) = 1.275g Cu(s)*(1 mol Cu(s) / 63.55g Cu(s) )
molCu (s) = 0.02mol Cu(s)


x mol Cu + 2y mol Cu = 0.02 mol Cu
Here I was able to cancel out the units:

mol Cu (x + 2y) = (0.02) mol Cu
x + 2y = 0.02

But the real problem is the first equation. How would I go about with canceling the units in the first equation?

because:

x mol CuO * (1 mol Cu /1 mol CuO)

y mol Cu2O * (2 mol Cu / 1 mol Cu2O)



You should have two equations rather than one:
\begin{aligned} \mathrm{CuO_{(s)}+{H_2}_{(g)}} & \xrightarrow{\quad\Delta\quad} \mathrm{Cu_{(s)}+{H_2O}_{(g)}}\\ \mathrm{{Cu_2O}_{(s)}+{H_2}_{(g)}} & \xrightarrow{\quad\Delta\quad} \mathrm{2\;Cu_{(s)}+{H_2O}_{(g)}} \end{aligned}

_________________
\begin{aligned} Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\ Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\ Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\ Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\ Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\ Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4 \end{aligned}
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A-R-Q
 Post subject:
PostPosted: Sat, 16 Jul 2011 17:31:46 UTC 
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You reply FAST. :)
Would you mind explaining to me what the triangle over the arrow means, and why it is relevant in this problem?
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outermeasure
 Post subject:
PostPosted: Sat, 16 Jul 2011 18:16:43 UTC 
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A-R-Q wrote:
You reply FAST. :)
Would you mind explaining to me what the triangle over the arrow means, and why it is relevant in this problem?


The \Delta in \xrightarrow{\quad\Delta\quad} means you need to heat it for the reaction to take place (at reasonable rate). It isn't actually relevant to the question, just for completeness sake (you heat up the oxide and then pass excess hydrogen through).

_________________
\begin{aligned} Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\ Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\ Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\ Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\ Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\ Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4 \end{aligned}
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A-R-Q
 Post subject:
PostPosted: Sat, 16 Jul 2011 21:22:22 UTC 
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Ok thanks, but I still require some assistance with Equation [1]. :|
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outermeasure
 Post subject:
PostPosted: Sun, 17 Jul 2011 10:54:28 UTC 
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A-R-Q wrote:
Ok thanks, but I still require some assistance with Equation [1]. :|


Suppose the original sample has x mol of CuO and y mol of Cu2O. Then their formula masses (using atomic masses 63.55 for copper and 16.00 for oxygen) give
1 mole of CuO = 79.55g
1 mole of Cu2O = 143.10g
So you get the first equation
(79.55)x+(143.10)y=1.512 ...[1]

_________________
\begin{aligned} Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\ Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\ Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\ Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\ Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\ Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4 \end{aligned}
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