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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: arzon138 on August 09, 2013, 01:59:19 AM

Title: Calculate the mass of each product that is expected. CO(g) + H2(g) --> CH3OH(l)
Post by: arzon138 on August 09, 2013, 01:59:19 AM

For the unbalanced chemical equation, suppose 10.0mg of each reactant is taken. Show by calculation which reactant is limiting. Calculate the mass of each product that is expected. CO(g) + H2(g) --> CH3OH(l)

Trying to figure out where i am gong wrong. I keep getting 5.41mg of CH3OH. The answers book shows 11.4mg of CH3OH. It does not show how to get to the answer however.

My working is as follows -

molar mass -
CO - 58.93
H2 - 2.016
CH3OH - 32.04

10Mg = 0.01g

0.01g Co X 1/58.93 = 1.69X10-4 Mol Co
0.01g H2 X 1/2.016 = 4.96X10-4 Mol H2

1.69X10-4 X 2/1 = 3.38X10-4 H2 - meaning Co is to be our limiting reactant

1.69X10-4 X 1/1 = 1.69X10-4 mol CH3OH

1.69X10-4 X 32.04/1 = 5.41 X 10-3g CH3OH X1000 = 5.41 mg....

Could someone please point out what i am doing wrong?

Thanks

Title: Re: Calculate the mass of each product that is expected. CO(g) + H2(g) --> CH3OH(l)
Post by: sjb on August 09, 2013, 02:08:20 AM

Quote from: arzon138 on August 09, 2013, 01:59:19 AM

For the unbalanced chemical equation, suppose 10.0mg of each reactant is taken. Show by calculation which reactant is limiting. Calculate the mass of each product that is expected. CO(g) + H2(g) --> CH3OH(l)

Trying to figure out where i am gong wrong. I keep getting 5.41mg of CH3OH. The answers book shows 11.4mg of CH3OH. It does not show how to get to the answer however.

My working is as follows -

molar mass -
CO - 58.93
H2 - 2.016
CH3OH - 32.04

10Mg = 0.01g

0.01g Co X 1/58.93 = 1.69X10-4 Mol Co
0.01g H2 X 1/2.016 = 4.96X10-4 Mol H2

1.69X10-4 X 2/1 = 3.38X10-4 H2 - meaning Co is to be our limiting reactant

1.69X10-4 X 1/1 = 1.69X10-4 mol CH3OH

1.69X10-4 X 32.04/1 = 5.41 X 10-3g CH3OH X1000 = 5.41 mg....

Could someone please point out what i am doing wrong?

Thanks

Check your molar mass of carbon monoxide.

And be careful, CO is not the same as Co, (and 10 Mg is not the same as 10 mg).

Title: Re: Calculate the mass of each product that is expected. CO(g) + H2(g) --> CH3OH(l)
Post by: arzon138 on August 09, 2013, 03:26:11 AM

Check your molar mass of carbon monoxide.

And be careful, CO is not the same as Co, (and 10 Mg is not the same as 10 mg).
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Thankyou. I think I've been at this for too long - can't believe I made such a silly mistake.