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3 years ago

Suppose that we combined 75 g of magnesium with 45 g of nitrogen and heat them up. Predict the product of

Chemistry

1 answer:

juin [17]3 years ago

6 0

Answer:

Mg is the limiting reactant

The theoretical yield is 105.2 g of Mg3N2

Explanation:

3Mg(s) + N2(g) -----> Mg3N2(s)

Number of moles of Mg = 75g/24g/mol = 3.125 moles

3 moles of Mg yields 1 mole of Mg3N2

3.125 moles of Mg yields 3.125 × 1/3 = 1.042 moles of Mg3N2

Number of moles of N2 = 45g/28g/mol = 1.607 moles

1 mole of N2 yields 1 mole of Mg3N2

1.607 moles of N2 yields 1.607 moles Mg3N2

Hence, Mg is the limiting reactant

The theoretical yield of Mg3N2 = 1.042 moles of Mg3N2 × 101 g/mol = 105.2 g of Mg3N2

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3 years ago

3. A 31.2-g piece of silver (s = 0.237 J/(g · °C)), initially at 277.2°C, is added to 185.8 g of a liquid, initially at 24.4°C,

VARVARA [1.3K]

Answer:

$Cp_{liquid}=2.54\frac{J}{g\°C}$

Explanation:

Hello,

In this case, since silver is initially hot as it cools down, the heat it loses is gained by the liquid, which can be thermodynamically represented by:

$Q_{Ag}=-Q_{liquid}$

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$m_{Ag}Cp_{Ag}(T_2-T_{Ag})=-m_{liquid}Cp_{liquid}(T_2-T_{liquid})$

Since no phase change is happening. Thus, solving for the heat capacity of the liquid we obtain:

$Cp_{liquid}=\frac{m_{Ag}Cp_{Ag}(T_2-T_{Ag})}{-m_{liquid}(T_2-T_{liquid})} \\\\Cp_{liquid}=\frac{31.2g*0.237\frac{J}{g\°C}*(28.3-227.2)\°C}{185.8g*(28.3-24.4)\°C}\\ \\Cp_{liquid}=2.54\frac{J}{g\°C}$

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3 years ago

When .080 moles of propane burn at STP, what volume of carbon dioxide is produced?

ahrayia [7]

Taking into account the reaction stoichiometry and the definition of STP, 5.4 L of carbon dioxide is produced.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

C₃H₈: 1 mole
O₂: 5 moles
CO₂: 3 moles
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<h3>Moles of CO₂ formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 1 mole of C₃H₈ form 3 moles of CO₂, 0.080 moles of C₃H₈ form how many moles of CO₂?

$amount of moles of CO_{2} =\frac{0.080 moles of C_{3} H_{8}x3 moles of CO_{2} }{1 mole of C_{3} H_{8} }$

<u><em>amount of moles of CO₂= 0.24 moles</em></u>

<h3>Definition of STP condition</h3>

The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases. And in these conditions 1 mole of any gas occupies an approximate volume of 22.4 liters.

<h3>Volume of CO₂ produced</h3>

You can apply the following rule of three: If by definition of STP 1 mole of CO₂ occupies 22.4 L, 0.24 moles of CO₂ how much volume does it occupy?

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