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3 years ago

Pre-lab Assignment

Chemistry

1 answer:

alina1380 [7]3 years ago

3 0

Answer:

1. d. Person A rounded their measurement

2) a. 105.

b. 76.4

c. 0.04

d. 1.988

Explanation:

1. Measuring an object is done with the least precise one. Person A recorded 1.5cm while Person B recorded 1.50cm of the same object. This measurement differs in the sense that Person A rounded up their measurement to 2 significant figures (1.5) while Person B recorded their answer as three significant figures (1.50).

2. During multiplication and division, the result is written as the measurement in the lowest significant figure.

a. 7.00 x 15.00 = 105. Calculator showed 105.00, which is 5 s.f (two zeros after decimal point are significant). The least significant figure is 3, hence 105.00 is rounded up to 105.

b. 40.2.1.901 = 76.4; The calculator answer of this operation is 76.420200 but since the lowest significant figure is 3. We round up our answer to 3 s.f i.e. 76.4

c. (0.003)(13.2) = 0.04. For this operation, calculator answer is 0.0396, however, the lowest significant figure in the question is 1 i.e. 0.003. Note that the two zeros before 3 are insignificant. Hence, our answer will be rounded up to 1 s.f i.e. 0.04

d. 1.590+ 0.3975 = 1.988. the calculator answer to this operation is 1.9875. However, the least significant figure in the question is 4 i.e. 1.590. hence, the answer will be rounded up to 4 s.f which is 1.988.

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Answer:

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When 100.g Mg3N2 reacts with 75.0 g H2O, what is the maximum theoretical yield of NH3?

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Answer : The correct option is, 23.6 g

Explanation : Given,

Mass of $Mg_3N_2$ = 100.0 g

Mass of $H_2O$ = 75.0 g

Molar mass of $Mg_3N_2$ = 101 g/mol

Molar mass of $H_2O$ = 18 g/mol

First we have to calculate the moles of $Mg_3N_2$ and $H_2O$ .

$\text{Moles of }Mg_3N_2=\frac{\text{Given mass }Mg_3N_2}{\text{Molar mass }Mg_3N_2}$

$\text{Moles of }Mg_3N_2=\frac{100.0g}{101g/mol}=0.990mol$

and,

$\text{Moles of }H_2O=\frac{\text{Given mass }H_2O}{\text{Molar mass }H_2O}$

$\text{Moles of }H_2O=\frac{75.0g}{18g/mol}=4.17mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$Mg_3N_2(s)+6H_2O(l)\rightarrow 3Mg(OH)_2(aq)+2NH_3(g)$

From the balanced reaction we conclude that

As, 6 moles of $H_2O$ react with 1 mole of $Mg_3N_2$

So, 4.17 moles of $H_2O$ react with $\frac{4.17}{6}=0.695$ moles of $Mg_3N_2$

From this we conclude that, $Mg_3N_2$ is an excess reagent because the given moles are greater than the required moles and $H_2O$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NH_3$

From the reaction, we conclude that

As, 6 moles of $H_2O$ react to give 2 moles of $NH_3$

So, 4.17 moles of $H_2O$ react to give $\frac{2}{6}\times 4.17=1.39$ mole of $NH_3$

Now we have to calculate the mass of $NH_3$

$\text{ Mass of }NH_3=\text{ Moles of }NH_3\times \text{ Molar mass of }NH_3$

Molar mass of $NH_3$ = 17 g/mole

$\text{ Mass of }NH_3=(1.39moles)\times (17g/mole)=23.6g$

Therefore, the maximum theoretical yield of $NH_3$ is, 23.6 grams.

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