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Lady_Fox [76]
3 years ago
13

Antacids, such as Alka-Seltzer, use the reaction of sodium bicarbonate with citric acid in water solution to produce a fizz as f

ollows: 3NaHCO3 + C6H8O7 → 3CO2 + 3H2O + Na3C6H5O7 If 4.11 g of the citric acid (C6H8O7, MW = 192 g/mol) react with excess sodium bicarbonate (NaHCO3), how many grams of carbon dioxide (CO2, MW = 44 g/mol) are formed as the solution fizzes?
Chemistry
1 answer:
Alex17521 [72]3 years ago
8 0

Answer:

2.8248 g

Explanation:

First, consider the balanced chemical equation:

3NaHCO_{3} +C_{6}H_{8}O_{7} --->3CO_{2} +3H_{2}O +Na_{3}C_{6}H_{5}O_{7}\\

Then we calculate the number of moles in 4.11 g of citric acid:

n(citric acid)=\frac{4.11g}{192g/mol}=0.0214mol

According to the balanced reaction, one mole of citric acid produces 3 moles of carbon dioxide. That's 3 times the number f moles of citric acid. So we will do the same with the available number of moles of citric acid.

so n(carbon dioxide) = 0.0214 mol*3=0.0642 mol

mass(carbon dioxide)= mass*molar mass=0.0642 mol* 44g/mol

                                                                   = 2.8248 g

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What is the kinetic evidence for SN1 reaction
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Answer:

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3 years ago
Manganese dioxide (MnO2(s), Hf = –520.0 kJ) reacts with aluminum to form aluminum oxide (AI2O3(s), Hf = –1699.8 kJ/mol) and mang
Temka [501]

Answer : The enthalpy of the reaction = -1839.6 KJ

Solution : Given,

\Delta (H_{f})_{MnO_{2}} = -520.0 KJ/mole

\Delta (H_{f})_{Al_{2}O_{3}} = -1699.8 KJ/mole

The balanced chemical reaction is,

3MnO_{2}(s)+4Al(s)\rightarrow 2Al_{2}O_{3}(s)+3Mn(s)

Formula used :

\Delta (H_{f})_{reaction}=\sum n(\Delta H_{f})_{product}-\sum n(\Delta H_{f})_{reactant}

\Delta (H_{f})_{reaction}=(2\times \Delta H_{Al_{2}O_{3}(s)}+3\times \Delta H_{Mn(s)} )-(3\times \Delta H_{MnO_{2}(s) }+4\times\Delta H_{Al}(s))

We know that the standard enthalpy of formation of the element is equal to Zero.

Therefore, the enthalpy of formation of (Mn) and (Al) is equal to zero.

Now, put all the values in above formula, we get

\Delta (H_{f})_{reaction}=[2moles\times (-1699.8 KJ/mole)}+3moles\times (0\text{ KJ/mole}})]-[(3moles\times(-520.0KJ/mole }+4moles\times(0\text{ KJ/mole})]

                        = (-3399.6) + (1560)

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4 years ago
a sample of sulfur dioxide occupies a volume of 652 mL at 40.0 C and 0.75 atm. What volume will the sulfur dioxide occupy at STP
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The volume that sulfur dioxide will occupy with a volume of 652 mL at 40.0°C and 0.75 atm is 0.019moles. Details about volume can be found below.

<h3>How to calculate volume?</h3>

The volume of a gas can be calculated using the following formula:

PV = nRT

  • P = pressure
  • V = volume
  • n = number of moles
  • R = gas law constant
  • T = temperature

0.75 × 0.652 = n × 0.0821 × 313

0.489 = 25.69n

n = 0.489/25.69

n = 0.019moles

Therefore, the volume that sulfur dioxide will occupy with a volume of 652 mL at 40.0°C and 0.75 atm is 0.019moles.

Learn more about volume at: brainly.com/question/1578538

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