Math / Algebra

Question $\mathrm{H}_{2}+\mathrm{CO}_{2} \rightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{CO}$
If 48.87 liters of $\mathrm{H}_{2}$ are reacted, how many liters of $\mathrm{CO}_{2}$ will also be reacted at STP? $\square$ liters of $\mathrm{CO}_{2}$

Studdy Solution

STEP 1

What is this asking? If we use up a certain amount of hydrogen to make water and carbon monoxide, how much carbon dioxide gets used up too? Watch out! Make sure to balance the equation first!
Don't forget about Avogadro's Law, which says equal volumes of gases at the same temperature and pressure have the same number of molecules.

STEP 2

1. Balance the Equation
2. Calculate the Volume of Carbon Dioxide

STEP 3

Alright, let's balance this chemical equation!
We've got $H_2 + CO_2 \rightarrow H_2O + CO$ .

STEP 4

Notice that there are two hydrogen atoms on both sides, one carbon atom on both sides, but three oxygen atoms on the right and only two on the left.
To balance the oxygen, we need to find the least common multiple of 2 and 3, which is 6.

STEP 5

So, if we put a 2 in front of $H_2O$ , we get $H_2 + CO_2 \rightarrow 2H_2O + CO$ .
Now we have four hydrogen atoms on the right, so we need to put a 2 in front of $H_2$ on the left side.
This gives us $2H_2 + CO_2 \rightarrow 2H_2O + CO$ .

STEP 6

Now we have four hydrogen atoms on both sides, one carbon atom on both sides, and two oxygen atoms on both sides. Balanced! Our balanced equation is $2H_2 + CO_2 \rightarrow 2H_2O + CO$ .

STEP 7

Now, we know that 48.87 liters of $H_2$ are reacted.
Because the gases are at STP, we can use the coefficients in the balanced equation to figure out the volume of $CO_2$ reacted.

STEP 8

The balanced equation tells us that 2 moles of $H_2$ react with 1 mole of $CO_2$ .
Thanks to Avogadro, we know that equal volumes of gases at STP contain the same number of moles.
So, the volume ratio is the same as the mole ratio!

STEP 9

If 2 liters of $H_2$ react with 1 liter of $CO_2$ , then 1 liter of $H_2$ reacts with $\frac{1}{2}$ liter of $CO_2$ .

STEP 10

We have 48.87 liters of $H_2$ , so we multiply that by $\frac{1}{2}$ to find the volume of $CO_2$ reacted:
$48.87 \cdot \frac{1}{2} = 24.435$

STEP 11

Therefore, 24.435 liters of $CO_2$ will be reacted.

STEP 12

24.435 liters of $CO_2$ will be reacted.

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Question $\mathrm{H}_{2}+\mathrm{CO}_{2} \rightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{CO}$
If 48.87 liters of $\mathrm{H}_{2}$ are reacted, how many liters of $\mathrm{CO}_{2}$ will also be reacted at STP? $\square$ liters of $\mathrm{CO}_{2}$

Studdy Solution

STEP 1

STEP 2

1. Balance the Equation
2. Calculate the Volume of Carbon Dioxide

STEP 3

Alright, let's balance this chemical equation!
We've got $H_2 + CO_2 \rightarrow H_2O + CO$ .

STEP 4

Notice that there are two hydrogen atoms on both sides, one carbon atom on both sides, but three oxygen atoms on the right and only two on the left.
To balance the oxygen, we need to find the least common multiple of 2 and 3, which is 6.

STEP 5

So, if we put a 2 in front of $H_2O$ , we get $H_2 + CO_2 \rightarrow 2H_2O + CO$ .
Now we have four hydrogen atoms on the right, so we need to put a 2 in front of $H_2$ on the left side.
This gives us $2H_2 + CO_2 \rightarrow 2H_2O + CO$ .

STEP 6

Now we have four hydrogen atoms on both sides, one carbon atom on both sides, and two oxygen atoms on both sides. Balanced! Our balanced equation is $2H_2 + CO_2 \rightarrow 2H_2O + CO$ .

STEP 7

Now, we know that 48.87 liters of $H_2$ are reacted.
Because the gases are at STP, we can use the coefficients in the balanced equation to figure out the volume of $CO_2$ reacted.

STEP 8

The balanced equation tells us that 2 moles of $H_2$ react with 1 mole of $CO_2$ .
Thanks to Avogadro, we know that equal volumes of gases at STP contain the same number of moles.
So, the volume ratio is the same as the mole ratio!

STEP 9

If 2 liters of $H_2$ react with 1 liter of $CO_2$ , then 1 liter of $H_2$ reacts with $\frac{1}{2}$ liter of $CO_2$ .

STEP 10

We have 48.87 liters of $H_2$ , so we multiply that by $\frac{1}{2}$ to find the volume of $CO_2$ reacted:
$48.87 \cdot \frac{1}{2} = 24.435$

STEP 11

Therefore, 24.435 liters of $CO_2$ will be reacted.

STEP 12

24.435 liters of $CO_2$ will be reacted.

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Studdy Solution

STEP 1

STEP 2

1. Balance the Equation2. Calculate the Volume of Carbon Dioxide

STEP 3

Alright, let's **balance** this chemical equation!We've got H2+CO2→H2O+COH_2 + CO_2 \rightarrow H_2O + COH2​+CO2​→H2​O+CO.

STEP 4

Notice that there are two hydrogen atoms on both sides, one carbon atom on both sides, but **three** oxygen atoms on the right and only **two** on the left.To balance the oxygen, we need to find the least common multiple of 2 and 3, which is **6**.

STEP 5

STEP 6

Now we have four hydrogen atoms on both sides, one carbon atom on both sides, and two oxygen atoms on both sides. **Balanced!** Our balanced equation is 2H2+CO2→2H2O+CO2H_2 + CO_2 \rightarrow 2H_2O + CO2H2​+CO2​→2H2​O+CO.

STEP 7

Now, we know that **48.87 liters** of H2H_2H2​ are reacted.Because the gases are at STP, we can use the **coefficients** in the balanced equation to figure out the volume of CO2CO_2CO2​ reacted.

STEP 8

The balanced equation tells us that **2 moles** of H2H_2H2​ react with **1 mole** of CO2CO_2CO2​.Thanks to Avogadro, we know that equal volumes of gases at STP contain the same number of moles.So, the **volume ratio** is the same as the **mole ratio**!

STEP 9

If 2 liters of H2H_2H2​ react with 1 liter of CO2CO_2CO2​, then 1 liter of H2H_2H2​ reacts with 12\frac{1}{2}21​ liter of CO2CO_2CO2​.

STEP 10

We have **48.87 liters** of H2H_2H2​, so we multiply that by 12\frac{1}{2}21​ to find the volume of CO2CO_2CO2​ reacted:48.87⋅12=24.43548.87 \cdot \frac{1}{2} = 24.43548.87⋅21​=24.435

STEP 11

Therefore, **24.435 liters** of CO2CO_2CO2​ will be reacted.

STEP 12

24.435 liters of CO2CO_2CO2​ will be reacted.

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Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. Balance the Equation
2. Calculate the Volume of Carbon Dioxide

Alright, let's balance this chemical equation!
We've got $H_2 + CO_2 \rightarrow H_2O + CO$ .

Notice that there are two hydrogen atoms on both sides, one carbon atom on both sides, but three oxygen atoms on the right and only two on the left.
To balance the oxygen, we need to find the least common multiple of 2 and 3, which is 6.

Now we have four hydrogen atoms on both sides, one carbon atom on both sides, and two oxygen atoms on both sides. Balanced! Our balanced equation is $2H_2 + CO_2 \rightarrow 2H_2O + CO$ .

Now, we know that 48.87 liters of $H_2$ are reacted.
Because the gases are at STP, we can use the coefficients in the balanced equation to figure out the volume of $CO_2$ reacted.

The balanced equation tells us that 2 moles of $H_2$ react with 1 mole of $CO_2$ .
Thanks to Avogadro, we know that equal volumes of gases at STP contain the same number of moles.
So, the volume ratio is the same as the mole ratio!

If 2 liters of $H_2$ react with 1 liter of $CO_2$ , then 1 liter of $H_2$ reacts with $\frac{1}{2}$ liter of $CO_2$ .

We have 48.87 liters of $H_2$ , so we multiply that by $\frac{1}{2}$ to find the volume of $CO_2$ reacted:
$48.87 \cdot \frac{1}{2} = 24.435$

Therefore, 24.435 liters of $CO_2$ will be reacted.

24.435 liters of $CO_2$ will be reacted.