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kumpel [21]
3 years ago
9

During a lab, you heat 1.04 g of a niso4 hydrate over a bunsen burner. after heating, the final mass of the dehydrated compound

is 0.61 g. determine the formula of the hydrate and also give the full name of the hydrate.
Chemistry
1 answer:
slamgirl [31]3 years ago
5 0
The difference between the initial mass (1.04 g) and the final mass (0.61g) is the mass of water present in the sample:

1.04 g - 0.61 g = 0.43 g

Divide by the molar mass of water: 18.01 g / mol to obtain the number of moles of water:

0.43 g / 18.01 g / mol = 0.0239 moles

Calculate the number of moles of dehydrated NiSO4

Number of moles = mass in grams / molar mass

molar mass of NiSO4 = 154.75 g/mol

number of moles NiSO4 = 0.61 g / 154.75 g/mol = 0.00394 moles

Now determine the molar ratio water to NiSO4: 0.0239 / 0.00394 = 6.07 ≈ 6

Then, the formula of the hydrate product is NiSO4 · 6 H2O.

Use the prefix hexa (6) to name the compound: Nickel(II) sulfate hexadydrate
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How many grams of ammonia (NH3) can be produced by the synthesis of excess hydrogen gas (H2) and 253.8 grams of nitrogen gas (N2
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Answer:

308.2 g of NH₃.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

3H₂ + N₂ —> 2NH₃

Next, we shall determine the mass of N₂ that reacted and the mass of NH₃ produced from the balanced equation. This can be obtained as follow:

Molar mass of N₂ = 2 × 14 = 28 g/mol

Mass of N₂ from the balanced equation = 1 × 28 = 28 g

Molar mass of NH₃ = 14 + (3×1)

= 14 + 3 = 17 g/mol

Mass of NH₃ from the balanced equation = 2 × 17 = 34 g

Summary:

From the balanced equation above,

28 g of N₂ reacted to produce 34 g of NH₃.

Finally, we shall determine the mass of NH₃ produced by the reaction of 253.8 g of N₂. This can be obtained as illustrated below:

From the balanced equation above,

28 g of N₂ reacted to produce 34 g of NH₃.

Therefore, 253.8 g of N₂ will react to produce = (253.8 × 34)/28 = 308.2 g of NH₃.

Thus, 308.2 g of NH₃ were obtained from the reaction.

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