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3 years ago

CuSO4.5H2O(k) ısı CuSO4(k) + 5 H2O(g) eşitliğine göre bakır sülfatın kristal suyu miktarı ve kaba formülü hesaplanacaktır.

Chemistry

1 answer:

baherus [9]3 years ago

3 0

Answer:

159.609 g/mol

Explanation:

According to the CuSO4.5H2O (k) heat CuSO4 (k) + 5 H2O (g) equation, the crystal water amount of copper sulfate and its rough formula will be calculated.

Weight of copper sulfate containing crystal water = m1 = 249.62… g

Weight of copper sulfate without crystal water weighed = m2 = 159.62 g

Accordingly, calculate the x and y values in the molecular formula of copper sulfate (xCuSO4.yH2O).

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I NEED HELP PLEASE!!!! CHEMISTRY QUESTION: If 38 g of Li3P and 15 grams of Al2O3 are reacted, what total mass of products will r

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Answer:

21.5 g.

Explanation:

Hello!

In this case, since the reaction between the given compounds is:

$2Li_3P+Al_2O_3\rightarrow 3Li_2O+2AlP$

We can see that according to the law of conservation of mass, which states that matter is neither created nor destroyed during a chemical reaction, the total mass of products equals the total mass of reactants based on the stoichiometric proportions; in such a way, we first need to compute the reacted moles of Li3P as shown below:

$n_{Li_3P}^{reacted}=38gLi_3P*\frac{1molLi_3P}{51.8gLi_3P}=0.73molLi_3P$

Now, the moles of Li3P consumed by 15 g of Al2O3:

$n_{Li_3P}^{consumed \ by \ Al_2O_3}=15gAl_2O_3*\frac{1molAl_2O_3}{101.96gAl_2O_3} *\frac{2molLi_3P}{1molAl_2O_3} =0.29molLi_3P$

Thus, we infer that just 0.29 moles of 0.73 react to form products; which means that the mass of formed products is:

$m_{Li_2O}=0.29molLi_3P*\frac{3molLi_2O}{2molLi_3P} *\frac{29.88gLi_2O}{1molLi_2O} =13gLi_2O\\\\m_{AlP}=0.29molLi_3P*\frac{2molAlP}{2molLi_3P} *\frac{57.95gAlP}{1molAlP} =8.5gAlP$

Therefore, the total mass of products is:

$m_{products}=13g+8.5g\\\\m_{products}=21.5g$

Which is not the same to the reactants (53 g) because there is an excess of Li₃P.

Best Regards!

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3 years ago