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Art [367]
3 years ago
11

A 3.82-g sample of magnesium nitride is reacted with 7.73 g of water. mg3n2 (s) + 3 h2o (â) â 2 nh3 (g) + 3 mgo (s) the yield of

mgo is 3.60 g. what is the percent yield in the reaction?
Chemistry
1 answer:
horrorfan [7]3 years ago
5 0
The reaction is properly written as

Mg₃N₂ (s) + 3 H₂O (l) --> 2 NH₃<span> (g) + 3 MgO (s) 

Molar mass of Mg</span>₃N₂ = 100.95 g/mol
Molar mass of H₂O = 18 g/mol
Molar mass of MgO = 40.3 g/mol

Moles Mg₃N₂: 3.82/100.95 = 0.0378
Moles H₂O: 7.73/18 = 0.429 

Theo H₂O required for available Mg₃N₂: 0.0378*3/1 = 0.1134 mol
Hence, the limiting reactant is Mg₃N₂.
Thus,

Theoretical Yield = 0.0378 mol Mg₃N₂ * 3 mol MgO/Mg₃N₂ * 40.3 g/mol
Theo Yield = 4.57 g

Percent Yield = Actual Yield/Theo Yield * 100
Percent Yield = 3.60 g/4.57 g * 100 =<em> 78.77%</em>
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When 2.5 grams react, the will produce how many grams of reactant in a complete chemical reaction?
const2013 [10]

Answer:

2.5g

Explanation:

When the reaction goes into completion, they will produce 2.5g. This is complement the law of conservation of mass.

According to the law of conservation of mass "in a chemical reaction, matter is neither created nor destroyed but transformed from one form to another".

  • The mass of reactants and products in a chemical reaction must be the same.
  • There is no change in mass in moving from reactant to product
  • So, if we start with 2.5g of reactants, we must end with 2.5g of products.
5 0
2 years ago
Analysis of skunk spray yields a molecule with 44.77% c, 7.46% h and 47.76% s by mass. what is the empirical formula for this mo
stiv31 [10]
Suppose we have 100 gr of the substance. Then by weight, it would contain 44.77 gr of C, 7.46 gr of H and 47.76 gr of S. We need to look up the atomic weights of these atoms; M_H=1, M_C=12, M_S=32. The following formula holds (where n are the moles of the substance, M its molecular mass and m its mass): n=m/M. Substituting the known quantities for each element, we get that the substance has 3.73 moles of C, 7.46 moles of H and 1.49 moles of S. In the empirical formula for the molecule, all atoms appear an integer amout of times. Hence, for every mole of Sulfur, we have 2.5 moles of C and 5 moles of H (by taking the moles ratios). Thus, for every 2 moles of sulfur, we have 5 moles of C and 10 moles of H. Now that all the coefficients are integer, we have arrived at an empirical formula for the skunk spray agent: C_5H_{10}S_2
4 0
3 years ago
How does mass affect a star's life cycle?
Andre45 [30]

Answer:

<h2>A star's life cycle is determined by its mass. The larger its mass, the shorter its life cycle. A star's mass is determined by the amount of matter that is available in its nebula, the giant cloud of gas and dust from which it was born. ... As the gas spins faster, it heats up and becomes as a protostar.</h2>

Explanation:

<h2>Read this and then choose your options ✍️✍️</h2>
3 0
3 years ago
When 412.5 g of calcium carbonate react with 521.9 g of aluminum fluoride, how many
LekaFEV [45]
Answer
321.8 g CaF2
321.5 g Al2(CO3)3
5 0
3 years ago
A sample of CaCO3 (molar mass 100. g) was reported as being 30. percent Ca. Assuming no calcium was present in any impurities, c
natka813 [3]

Answer:

Approximately 75%.

Explanation:

Look up the relative atomic mass of Ca on a modern periodic table:

  • Ca: 40.078.

There are one mole of Ca atoms in each mole of CaCO₃ formula unit.

  • The mass of one mole of CaCO₃ is the same as the molar mass of this compound: \rm 100\; g.
  • The mass of one mole of Ca atoms is (numerically) the same as the relative atomic mass of this element: \rm 40.078\; g.

Calculate the mass ratio of Ca in a pure sample of CaCO₃:

\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} = \frac{40.078}{100} \approx \frac{2}{5}.

Let the mass of the sample be 100 g. This sample of CaCO₃ contains 30% Ca by mass. In that 100 grams of this sample, there would be \rm 30 \% \times 100\; g = 30\; g of Ca atoms. Assuming that the impurity does not contain any Ca. In other words, all these Ca atoms belong to CaCO₃. Apply the ratio \displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} \approx \frac{2}{5}:

\begin{aligned} m\left(\mathrm{CaCO_3}\right) &= m(\mathrm{Ca})\left/\frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)}\right. \cr &\approx 30\; \rm g \left/ \frac{2}{5}\right. \cr &= 75\; \rm g \end{aligned}.

In other words, by these assumptions, 100 grams of this sample would contain 75 grams of CaCO₃. The percentage mass of CaCO₃ in this sample would thus be equal to:

\displaystyle 100\%\times \frac{m\left(\mathrm{CaCO_3}\right)}{m(\text{sample})} = \frac{75}{100} = 75\%.

3 0
3 years ago
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