[arzon138]

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

[sjb]

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).

[arzon138]

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).
[/quote]

Thankyou. I think I've been at this for too long - can't believe I made such a silly mistake.