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zvonat [6]
4 years ago
5

Calculate how many moles of NH3 form when each quantity of reactant completely reacts according to the equation:

Chemistry
1 answer:
sergiy2304 [10]4 years ago
7 0
The balanced chemical reaction is:

<span>3N2H4(l)→4NH3(g)+N2(g)
</span>
The amounts given for the N2H4 reactant will be the starting point for our calculations.

2.6mol N2H4 ( 4 mol NH3 / 3 mol N2H4 ) = 3.47 mol NH3

4.05mol N2H4 ( 4 mol NH3 / 3 mol N2H4 ) = 5.4 mol NH3

63.8g N2H4  <span>( 4 mol NH3 / 3 mol N2H4 ) = 85.07 mol NH3</span>
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Which best describes how magnesium (Mg) and oxygen (O) bond?
Ksivusya [100]

Answer:

a

They form an ionic bond by exchanging two electrons.

Explanation:

Magnesium loses two electrons and oxygen gains these electrons so ionic bond is formed

8 0
3 years ago
If the elastic modulus of cobalt (co) is 200 gpa, and the elastic modulus of tungsten carbide (wc) is 700 gpa, calculate the upp
Dmitry_Shevchenko [17]
To determine the upper bond
Ec(u) = EmVm + EpVp
Em is the elastic modulus of cobalt.
E₁ is the elastic modulus of the particulate
Vm is the volume fraction of cobalt
Vp is the volume fraction of particulate
substitute
Ec(u) = 200 (Vm) + 700 (Vp)
To determine the lower bound
Ec (l) = EmEp/VmEp+ VpEm
Substitute
Ec (l) = 200(700)/Vm(700) + Vp (200)
Ec (l) = 1400/7Vm+2Vp
7 0
4 years ago
I need help with my chemistry homework please &amp; thank u
lisov135 [29]

Answer:

6) solid- dots close together

liquid- a little bit farther apart than solid

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5 0
3 years ago
Do Alkali Metal Compounds form precipitates? And Why?
Alexeev081 [22]

Answer:

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6 0
2 years ago
Find percent yield:
saveliy_v [14]

<u>Answer:</u> The percent yield of the reaction is 91.8 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

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

  • <u>For B_5H_9 :</u>

Given mass of B_5H_9 = 4.0 g

Molar mass of B_5H_9 = 63.12 g/mol

Putting values in equation 1, we get:

\text{Moles of }B_5H_9=\frac{4g}{63.12g/mol}=0.0634mol

  • <u>For oxygen gas:</u>

Given mass of oxygen gas = 10.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

\text{Moles of oxygen gas}=\frac{10g}{32g/mol}=0.3125mol

The chemical equation for the reaction of B_5H_9 and oxygen gas follows:

2B_5H_9+12O_2\rightarrow 5B_2O_3+9H_2O

By Stoichiometry of the reaction:

12 moles of oxygen gas reacts with 2 moles of B_2H_5

So, 0.3125 moles of oxygen gas will react with = \frac{2}{12}\times 0.3125=0.052mol of B_2H_5

As, given amount of B_2H_5 is more than the required amount. So, it is considered as an excess reagent.

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

By Stoichiometry of the reaction:

12 moles of oxygen gas produces 5 moles of B_2O_3

So, 0.3125 moles of oxygen gas will produce = \frac{5}{12}\times 0.3125=0.130moles of water

Now, calculating the mass of B_2O_3 from equation 1, we get:

Molar mass of B_2O_3 = 69.93 g/mol

Moles of B_2O_3 = 0.130 moles

Putting values in equation 1, we get:

0.130mol=\frac{\text{Mass of }B_2O_3}{69.63g/mol}\\\\\text{Mass of }B_2O_3=(0.130mol\times 69.63g/mol)=9.052g

To calculate the percentage yield of B_2O_3, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of B_2O_3 = 8.32 g

Theoretical yield of B_2O_3 = 9.052 g

Putting values in above equation, we get:

\%\text{ yield of }B_2O_3=\frac{8.32g}{9.052g}\times 100\\\\\% \text{yield of }B_2O_3=91.8\%

Hence, the percent yield of the reaction is 91.8 %

6 0
4 years ago
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