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ArbitrLikvidat [17]
2 years ago
14

A chemical company makes pure silver by reacting silver nitrate with zinc. the company needs to make 800 grams of pure silver fo

r a client. they have 500 grams of zinc and 1500 grams of silver nitrate. will they be able to make enough silver to fill the order? (you may have to balance the equation.) zn agno3 -> zn(no3)2 ag
Chemistry
1 answer:
Ira Lisetskai [31]2 years ago
7 0

Yes, they will be able to make enough silver to fill the order. They possess more reactants than are necessary.

Zinc reduces silver nitrate to silver metal according to the following equation.

2AgNO₃ + Zn → 2Ag + Zn(NO₃)₂

From the equation

2 moles of AgNO₃ produce 2 moles of silver.

If we consider the zinc to be used number of moles=mass/RAM

RAM of zinc =65.38

No. of moles=500/65.38

=7.6476moles

To produce 800 grams of silver they require:

800/107.868 moles=7.4165 moles of silver nitrate since the ratio of silver nitrate to silver produced is 1:1

The number of moles of silver nitrate available=1500/169.872

=8.83 moles.

As the amount of reactants available is more than the required the company will make it in producing the 800 grams of silver required.

<h3>More about silver nitrate</h3>

An inorganic substance with the chemical formula AgNO3 is called silver nitrate. It is coordinated in a trigonal planar structure when it is solid. It is frequently employed as a precursor to other compounds that include silver. As a staining agent for protein visibility in PAGE gels and scanning electron microscopy, it is utilized in the production of photographic films as well as in lab settings.

To view more questions about silver nitrate, refer to:

brainly.com/question/8071605

#SPJ4

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RideAnS [48]

Answer:

TS (total solid, mg/L) = 57,174 mg/L

VS (volatile solid, mg/L) = 24,088 mg/L

Explanation:

First step to solve this problem is to know data and questions:

Data:

Sample volume = 50 mL

Sample weight = 50.1 g

P1 (weight empty crucible) = 17.1234 g

P2 (weight after drying a 103º in oven) = 19.9821  g

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Questions:

TS = ?  

VS = ?

Formula:

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TS =\frac{(P2-P1)*\frac{1,000 mg}{1g} }{Sample volumen (L)}

VS = \frac{(P2-P3)*\frac{1,000 mg}{1g} }{Sample volume (L)}

We have to transform mL to L so we will divide mL by 1,000 the sample volume:

Sample Volumen (L) = 50 mL * \frac{1L}{1,000 mL} = 0.05 L

In the formula the value of 1,000 results by the convertion factor to transform grams to miligrams (we have to multiplie by 1,000)

Now we need to replace data on previous formulas and we will get TS and VS expressed in mg/L:

TS = \frac{(19.9821 g-17.1234g)*\frac{1,000 mg}{1g} }{0.05L} = 57,174mg/L

VS = \frac{(19.9821g-18.7777g)*\frac{1,000mg}{1g} }{0.05L}=24,088 mg/L

We divide TS and VS formula by sample volume because the exercise is asking us to express the results in mg/L.

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Answer:

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Explanation:

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