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Anton [14]
2 years ago
12

¿Cuántos moles hay en 350 g de NaNO3?

Chemistry
1 answer:
olga55 [171]2 years ago
4 0

Answer:

3 moles

Explanation:

So, approximately 3 moles of NaNO3 can be obtained, by reacting 253 grams of Na2CrO4. Also, the number of moles is a dimensionless quantity.

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What is 5 divided by 2.7​
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2 years ago
If you begin with 2.7 g Al and 4.05 g Cl2, what mass of AlCl3 can be produced?
Tresset [83]
<span>atomic weights: Al = 26.98, Cl = 35.45 In this reaction; 2Al = 53.96 and 3Cl2 = 212.7 Ratio of Al:Cl = 53.96/212.7 = 0.2537 that is approximately four times the mass Cl is needed. Step 2: (a) Ratio of Al:Cl = 2.70/4.05 = 0.6667 since the ratio is greater than 0.2537 the divisor which is Cl is not big enough to give a smaller ratio equal to 0.2537. so Cl is limiting (b)since Cl is the limiting reactant 4.05g will be used to determine the mass of AlCl3 that can be produced. From Step 1: 212.7g of Cl will produce 266.66g AlCl3 212.7g = 266.66g 4.05g = x x = 5.08g of AlCl3 can be produced (c) Al:Cl = 0.2537 Al:Cl = Al:4.05 = 0.2537 mass of Al used in reaction = 4.05 x 0.2537 = 1.027g Excess reactant = 2.70 - 1.027 = 1.67g King Leo · 9 years ago</span>
8 0
3 years ago
The addition of hydrochloric acid to a silver nitrate solution precipitates silver chloride according to the reaction:
Nimfa-mama [501]

Answer:

Enthalpy change for the reaction is -67716 J/mol.

Explanation:

Number of moles of AgNO_{3} in 50.0 mL of 0.100 M of AgNO_{3}

= Number of moles of HCl in 50.0 mL of 0.100 M of HCl

= \frac{0.100}{1000}\times 50.0 moles

= 0.00500 moles

According to balanced equation, 1 mol of AgNO_{3} reacts with 1 mol of HCl to form 1 mol of AgCl.

So, 0.00500 moles of AgNO_{3} react with 0.00500 moles of HCl to form 0.00500 moles of AgCl

Total volume of solution = (50.0+50.0) mL = 100.0 mL

So, mass of solution = (100.0\times 1.00) g = 100 g

Enthalpy change for the reaction = -(heat released during reaction)/(number of moles of AgCl formed)

= \frac{-m_{solution}\times C_{solution}\times \Delta T_{solution}}{0.00500mol}

= \frac{-100g\times 4.18\frac{J}{g.^{0}\textrm{C}}\times [24.21-23.40]^{0}\textrm{C}}{0.00500mol}

= -67716 J/mol

[m = mass, c = specific heat capacity, \Delta T = change in temperature and negative sign is included as it is an exothermic reaction]

4 0
2 years ago
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