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3 years ago

Use the unbalanced equation NH3+O2=NO+H2O to find the mole ratio between NO and H2O

Chemistry

1 answer:

Leona [35]3 years ago

5 0

Answer:

For every 4 moles of NO created, 6 moles of H2O are created so the ratio is 4:6

Explanation:

You just need to balance the equation.

NH3 + O2 -> NO + H2O

1. I started with hydrogen; there's 3 on the left and 2 on the right. Multiply them together to find a number they both go into (3×2=6, but in this case 6 hydrogen on each side does not work so I doubled it so there is 12 hydrogen on each side).

This will bring you to this:

4NH3 + O2 -> NO + 6H2O

2. Now get equal amounts of nitrogen on each side. There's 4 nitrogen on the left side, and 1 on the right. Multiply the right by 4. Then you will have this:

4NH3 + O2 -> 4NO + 6H2O

3. Last thing you need to do is have the same amount of oxygen on both sides. On the left you have 2 and on the right you have 10. Get the left to 10 by multiplying it by 5.

Balanced: 4NH3 + 5O2 -> 4NO + 6H2O

In word form, for every reaction between 4 moles of ammonia and 5 moles of oxygen, 4 moles of nitric oxide and 6 moles of water will be created.

I hope this helps!

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A 32.2 g iron rod, initially at 21.9 C, is submerged into an unknown mass of water at 63.5 C. in an insulated container. The fin

lorasvet [3.4K]

Answer : The mass of the water in two significant figures is, $3.0\times 10^1g$

Explanation :

In this case the heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of iron metal = $0.45J/g^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of iron metal = 32.3 g

$m_2$ = mass of water = ?

$T_f$ = final temperature of mixture = $59.2^oC$

$T_1$ = initial temperature of iron metal = $21.9^oC$

$T_2$ = initial temperature of water = $63.5^oC$

Now put all the given values in the above formula, we get

$32.3g\times 0.45J/g^oC\times (59.2-21.9)^oC=-m_2\times 4.18J/g^oC\times (59.2-63.5)^oC$

$m_2=30.16g\approx 3.0\times 10^1g$

Therefore, the mass of the water in two significant figures is, $3.0\times 10^1g$

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3 years ago

Concert 15 cm3 to liters

musickatia [10]

The answer to your problem is 0.015 liters. I got the answer because to convert cubic centimeters into liters, you need to divide the cubic centimeters by 1000.

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2 years ago

What's the difference between a mole of propane and a gram of propane? <br>

kobusy [5.1K]

B. A gram would have a lot more molecules of propane than a mole

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3 years ago

How many molecules are in 7.62 L of CH4, at 87.5°C and 722 torr

pickupchik [31]

Answer: There are $1.469 \times 10^{23}$ molecules present in 7.62 L of $CH_4$ at $87.5^{o}C$ and 722 torr.

Explanation:

Given : Volume = 7.62 L

Temperature = $87.5^{o}C = (87.5 + 273) K = 360.5 K$

Pressure = 722 torr

1 torr = 0.00131579

Converting torr into atm as follows.

$722 torr = 722 torr \times \frac{0.00131579 atm}{1 torr}\\= 0.95 atm$

Therefore, using the ideal gas equation the number of moles are calculated as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

$PV = nRT\\0.95 atm \times 7.62 L = n \times 0.0821 L atm/mol K \times 360.5 K\\n = \frac{0.95 atm \times 7.62 L}{0.0821 L atm/mol K \times 360.5 K}\\= \frac{7.239}{29.59705}\\= 0.244 mol$

According to the mole concept, 1 mole of every substance contains $6.022 \times 10^{23}$ atoms. Hence, number of atoms or molecules present in 0.244 mol are calculated as follows.

$0.244 mol \times 6.022 \times 10^{23}\\= 1.469 \times 10^{23}$

Thus, we can conclude that there are $1.469 \times 10^{23}$ molecules present in 7.62 L of $CH_4$ at $87.5^{o}C$ and 722 torr.

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3 years ago

Need help on 17,19,20 please.

pentagon [3]

Answer:

c, maybe d, and I think b.

Explanation:

Im sorry if wrong

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