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TiliK225 [7]
3 years ago
11

Predict the product(s) of the following reaction. (Separate substances in a list with a comma. Include states-of-matter under th

e given conditions in your answer.) Cu(NO3)2 + Rb3PO4 ------>(H2O above the arrow)
Chemistry
1 answer:
kiruha [24]3 years ago
6 0

Answer:

Cu₃(PO₄)₂(s), RbNO₃(aq)

Explanation:

This reaction looks like a possible <em>double replacement reaction</em>, in which the metal ions have exchanged partners.

3Cu(NO₃)₂(aq) + 2Rb₃PO₄(aq) ⟶ Cu₃(PO₄)₂ + 6RbNO₃

You must recall the pertinent <em>solubility rules</em>:

1. Salts of Group 1 elements (e.g., Rb⁺) are soluble.

2. Salts containing nitrate ions (NO₃⁻) are soluble.

3. Most salts containing phosphate ions (PO₄³⁻) are insoluble

According to Rules 1 and 1, RbNO₃ is soluble.

According to Rule 3, Cu₃(PO₄)₂ is insoluble.

∴ 3Cu(NO₃)₂(aq) + 2Rb₃PO₄(aq) ⟶ Cu₃(PO₄)₂(s) + 6RbNO₃(aq)

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How do the chemical characteristics of carbon affect the characteristics of organic molecules?
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Gaseous butane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 13. g of butane is m
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<u>Answer:</u> The maximum amount of water that could be produced by the chemical reaction is 20.16 grams

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\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

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Given mass of butane = 13 g

Molar mass of butane = 58.12 g/mol

Putting values in equation 1, we get:

\text{Moles of butane}=\frac{13g}{58.12g/mol}=0.224mol

  • <u>For oxygen gas:</u>

Given mass of oxygen gas = 70.9 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

\text{Moles of oxygen gas}=\frac{70.9g}{32g/mol}=2.216mol

The chemical equation for the reaction of butane and oxygen gas follows:

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By Stoichiometry of the reaction:

2 moles of butane reacts with 13 moles of oxygen gas

So, 0.224 moles of butane will react with = \frac{13}{2}\times 0.224=1.456mol of oxygen gas

As, given amount of oxygen gas is more than the required amount. So, it is considered as an excess reagent.

Thus, butane is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of butane produces 10 moles of water

So, 0.224 moles of butane will produce = \frac{10}{2}\times 0.224=1.12moles of water

Now, calculating the mass of water from equation 1, we get:

Molar mass of water = 18 g/mol

Moles of water = 1.12 moles

Putting values in equation 1, we get:

1.12mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(1.12mol\times 18g/mol)=20.16g

Hence, the maximum amount of water that could be produced by the chemical reaction is 20.16 grams

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