Co(g) + 2 h2 --> ch3oh 2.50 g of hydrogen is reacted with 30.0 l of carbon monoxide at stp. 1. what is the limiting reactant? *hint: only list the element symbol* 2. what mass of ch3oh is produced? *hint: only list the grams* 3. how much excess is left over? *hint: only list the grams*

Answer 1 : The limiting reactant is, hydrogen gas,

Solution :

First we have to calculate the moles of hydrogen gas.

Now we have to calculate the volume of hydrogen gas.

As, 1 mole of gas contains 22.4 L volume of gas

So, 1.25 mole of hydrogen gas contains volume of hydrogen gas

Now we have to calculate the limiting and excess reactant.

The given balanced reaction is,

From the balanced reaction, we conclude that

As, 44.8 L of hydrogen gas react with 22.4 L of carbon monoxide gas

So, 28 L of hydrogen gas react with of carbon monoxide gas

The excess carbon monoxide = 30 L - 14 L = 16 L

Thus, the carbon monoxide is an excess reactant because it is present in excess amount and hydrogen gas is a limiting reactant because it is present in limited amount.

Answer 2 : The mass of is, 20 grams

Solution :

From the balance reaction, we conclude that

2 moles of hydrogen gas react to give 1 mole of

1.25 moles of hydrogen gas react to give mole of

Now we have to calculate the mass of

The mass of is, 20 grams

Answer 3 : The amount of excess reactant is, 20 grams

The excess reactant is carbon monoxide.

As, 22.4 L volume of carbon monoxide gas has 28 gram of carbon monoxide gas

So, 16 L volume of carbon monoxide gas has gram of carbon monoxide gas

Thus, the The amount of excess reactant is, 20 grams

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