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nexus9112 [7]
3 years ago
13

Sodium phosphate and calcium chloride react to form sodium chloride and calcium phosphate. If you have 379.4 grams of calcium ch

loride and an excess of sodium phosphate, how much calcium phosphate can you make? Use the periodic table and the polyatomic ion resource.
A. 353.5 g
B. 379.4 g
C. 402.2 g
D. 863.4 g
E. 1,060 g
Chemistry
1 answer:
GalinKa [24]3 years ago
8 0

Answer:- A. 353.5 g

Solution:- The balanced equation is:

2Na_3PO_4(aq)+3CaCl_2(aq)\rightarrow 6NaCl(aq)+Ca_3(PO_4)_2(s)

From this equation, there is 3:1 mol ratio between calcium chloride and calcium phosphate.

Grams of calcium chloride are converted to moles and these moles are multiplied by the mol ratio to get the moles of calcium phosphate. Finally the moles are multiplied by molar mass to get the grams of calcium phosphate.

Molar mass of CaCl_2 = 40.078+2(35.45) = 110.978 g per mol

Molar mass of Ca_3(PO_4)_2 = 3(40.078)+2(30.974)+8(16.00)

= 120.234+61.948+128.00 = 310.182 g per mol

let's make the set up using dimensional analysis.

379.4gCaCl_2(\frac{1molCaCl_2}{110.978gCaCl_2})(\frac{1molCa_3(PO_4)_2}{3molCaCl_2})(\frac{310.182gCa_3(PO_4)_2}{1molCa_3(PO_4)_2})

= 353.5gCa_3(PO_4)_2

Hence, 353.5 g of calcium phosphate will form and so the correct option is A.

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Which is a common way for a scientific calculator to show the number below? <br> 8.1 102
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Answer: 8.1e+2 or 8.1E+2 (in the scientific notation mode), based on that the number given is 8.1 × 10².

Explanation


Scientific calculators use the letter e or E to show the numbers in scientific notation mode.

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6.022 × 10²³           6.022 e+23
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2 years ago
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
maksim [4K]

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!

7 0
3 years ago
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