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Arlecino [84]
3 years ago
5

Gaseous methane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water. Suppose 0.802 g of methane i

s mixed with 1.9 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to significant digits.
Chemistry
1 answer:
Kipish [7]3 years ago
8 0

Answer:

1.07g

Explanation:

Step 1:

We will begin by writing the balanced equation for the reaction. This is given below:

CH4 + 2O2 —> CO2 + 2H2O

Step 2:

Determination of the masses of CH4 and O2 that reacted and the mass of H2O produced from the balanced equation. This is illustrated below:

Molar Mass of CH4 = 12 + (4x1) = 12 + 4 = 16g/mol

Molar Mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 2 x 32 = 64g

Molar Mass of H2O = (2x1) + 16 = 2 + 16 = 18g/mol

Mass of H2O from the balanced equation = 2 x 18 = 36g

Summary:

From the balanced equation above,

16g of CH4 reacted with 64g of O2 to produce 36g of H2O.

Step 3:

Determination of the limiting reactant.

We need to know which of the reactant is limiting the reaction in order to obtain the maximum mass of water.

This is illustrated below:

From the balanced equation above,

16g of CH4 reacted with 64g of O2.

Therefore, 0.802g of CH4 will react with = (0.802 x 64)/16 = 3.21g of O2.

From the above calculations, a higher mass of O2 is needed to react with 0.802g of CH4. Therefore, O2 is the limiting reactant.

Step 4:

Determination of the mass of H2O produced from the reaction.

To obtain the maximum mass of H2O produced, the limiting reactant will be used because it will generate the maximum yield of the product.

From the balanced equation above,

64g of O2 produce 36g of H2O.

Therefore, 1.9g of O2 will produce = (1.9 x 36)/64 = 1.07g of H2O.

The maximum mass of water (H2O) produced by the reaction is 1.07g

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(part 1 of 3) Copper reacts with silver nitrate through a single replacement. If 1.29 g of silver are produced from the reaction
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Answer:

See explanation.

Explanation:

Hello there!

In this case, according to the described chemical reaction, we first write the corresponding equation to obtain:

Cu+2AgNO_3\rightarrow 2Ag+Cu(NO_3)_2

Thus, we proceed as follows:

Part 1 of 3: here, since the molar mass of silver and copper (II) nitrate are 107.87 and 187.55 g/mol respectively, and the mole ratio of the former to the latter is 2:1, we can set up the following stoichiometric expression:

m_{Cu(NO_3)_2}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{1molCu(NO_3)_2}{2molAg}*\frac{187.55gCu(NO_3)_2}{1molCu(NO_3)_2}   \\\\m_{Cu(NO_3)_2}=1.12gCu(NO_3)_2

Part 2 of 3: here, the molar mass of copper is 63.55 g/mol and the mole ratio of silver to copper is 2:1, the mass of the former that was used to start the reaction was:

m_{Cu}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{1molCu}{2molAg}*\frac{63.55gCu)_2}{1molCu}   \\\\m_{Cu}=0.380gCu

Part 3 of 3: here, the molar mass of silver nitrate is 169.87 g/mol and their mole ratio 2:2, thus, the mass of initial silver nitrate is:

m_{AgNO_3}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{2molAgNO_3}{2molAg}*\frac{169.87gAgNO_3}{1molAgNO_3}   \\\\m_{AgNO_3}=2.03gAgNO_3

Best regards!

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