How many grams of h2 are needed to produce 10.78 g of nh3?

1 answer:

6 0

Ammonia is formed by a reaction between hydrogen and nitrogen as shown by the equation below.
N2(g) + 3H2(g) = 2NH3(g)
1 mole of ammonia contains 17 g
Therefore 10.78 g of ammonia are equivalent to 10.78/17 = 0.6341 moles
The mole ratio of hydrogen to ammonia is 3 : 2
Therefore, moles of hydrogen used will be 0.6341 × 3/2 = 0.9512 moles
1 mole of hydrogen is equivalent to 2 g
Thus, the mas of hydrogen will be 0.9512 moles × 2 = 1.9023 g

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If a sample of sodium chloride with a mass of

alex41 [277]

Original molarity was 1.7 moles of NaCl

Final molarity was 0.36 moles of NaCl

Given Information:

Original (concentrated) solution: 25 g NaCl in a 250 mL solution, solve for molarity

Final (diluted) solution: More water is added to make the new total volume 1.2 liters, solve for the new molarity

1. Solve for the molarity of the original (concentrated) solution.

Molarity (M) = moles of solute (mol) / liters of solution (L)

Convert the given information to the appropriate units before plugging in and solving for molarity.

Molarity (M) = 0.43 mol NaCl solute / 0.250 L solution = 1.7 M NaCl (original solution)

2. Solve for the molarity of the final (diluted) solution.

Remember that the amount of solute remains constant in a dilution problem; it is just the total volume of the solution that changes due to the addition of solvent.

Molarity (M) = 0.43 mol NaCl solute / 1.2 L solution

Molarity (M) of the final solution = 0.36 M NaCl

I hope this helped:))

6 0

2 years ago

how many moles of sodium are needed to react with sulfuric acid to produce 3.75 moles of sodium sulfate according to the followi

anastassius [24]

Hello!
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You'll need to react 7,5 moles of Sodium with sulfuric acid to produce 3.75 moles of sodium sulfate
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First of all, you need to balance the reaction. The balanced reaction is shown below (ensuring that the Law of Conservation of Mass is met on both sides):

2Na + H₂SO₄ → Na₂SO₄ + H₂

Now, all that you have to do is to use molar equivalences in this reaction applying the coefficients to calculate the moles of Sodium that you'll need:

$molesNa=3,75moles Na_{2} SO_4* \frac{2 moles Na}{1 mol Na_{2} SO_4} =7,5 moles Na$

Have a nice day!

5 0

2 years ago

What does contact forces and non-contact forces mean?

Dmitrij [34]

A contact force is a force in which and object comes in contact with a another object. A non-contact force, is a force is an object applied by another body. (A good example of a non-contact force is gravity)

8 0

3 years ago

44.8% of a 250. mL acid solution is used in an experiment, what volume of acid (in mL) was used?

postnew [5]

Answer:

112 mL

Explanation:

The formula for percent by volume is

$\text{Percent by volume} = \dfrac{\text{Volume of solute}}{\text{Volume of solution}}\times 100 \, \%$

If you have 250 mL of a solution that is 44.8 % v/v,

$\begin{array}{rcl}44.8\, \% & = & \dfrac{\text{Volume of solute}}{\text{250 mL}}\times 100 \, \%\\\\44.8 \times \text{ 250 mL} & = & \text{Volume of solute} \times 100\\\text{Volume of solute} & = & \dfrac{44.8 \times 250\text{ mL}}{100}\\\\ & = & \textbf{112 mL}\\\end{array}$

7 0

3 years ago

An unknown compound has a percent composition of 52.10% potassium, 15.8% carbon, and 32.1% oxygen. The molar mass of the compoun

lianna [129]

To determine the empirical formula and the molecular formula of the compound, we assume a basis of the compound of 100 g. We do as follows:

Mass Moles
K 52.10 52.10/39.10 = 1.33 1.33/1.32 ≈ 1
C 15.8 15.8/12 = 1.32 1.32/1.32 ≈ 1
O 32.1 32.1 / 16 = 2.01 2.01/1.32 ≈ 1.5

The empirical formula would most likely be KCO.
The molecular formula would be K2C2O3.

4 0

3 years ago

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