Question

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

CO(g) +2H2--->CH3OH
2.50g H2*1mol/2g=1.25 mol H2
30.0L CO*1mol/22.4L=1.34 mol CO,
according to reaction 1 mol CO needs 2 mol H2,so 1.34 mol CO need 2.68 mol H2, so 1) limiting teactant is H2 (H)

2)1.25 mol CH3OH will be produced, 1.25 mol*32g/mol=40.0 g CH3OH
3) 1.25 mol H2 needs 0.625 g CO
1.34-0.625=0.715 g CO leftover

Answer 2

Answer 1 : The limiting reactant is, hydrogen gas,
`H_2`

Solution :

First we have to calculate the moles of hydrogen gas.

`\text{Moles of }H_2=\frac{\text{Mass of }H_2}{\text{Molar mass of }H_2}=(2.5g)/(2g/mole)=1.25moles`

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
`1.25* 22.4=28L` volume of hydrogen gas

Now we have to calculate the limiting and excess reactant.

The given balanced reaction is,

`CO(g)+2H_2(g)\rightarrow CH_3OH(l)`

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
`(22.4L)/(44.8L)* 28L=14L` 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
`CH_3OH` is, 20 grams

Solution :

From the balance reaction, we conclude that

2 moles of hydrogen gas react to give 1 mole of
`CH_3OH`

1.25 moles of hydrogen gas react to give
`(1.25)/(2)=0.625` mole of
`CH_3OH`

Now we have to calculate the mass of
`CH_3OH`

`\text{Mass of }CH_3OH=\text{Moles of }CH_3OH* \text{Molar mass of }CH_3OH`

`\text{Mass of }CH_3OH=(0.625mole)* (32g/mole)=20g`

The mass of
`CH_3OH` 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
`(28)/(22.4)* 16=20` gram of carbon monoxide gas

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