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attashe74 [19]
3 years ago
13

How many grams of CO2 evolved from a 1.205g sample that is 36% MgCO3 and 44% K2CO3 by mass?

Chemistry
1 answer:
emmasim [6.3K]3 years ago
7 0

Mass of CO₂ evolved : 0.108 g

<h3>Further explanation</h3>

Given

1.205g sample, 36% MgCO3 and 44% K2CO3

Required

mass of CO2

Solution

  • mass of MgCO₃ :

0.36 x 1.205 g=0.4338 g

mass C in MgCO₃(MW MgCO₃=84 g/mol,  Ar C = 12/gmol)

= (12/84) x 0.4338

= 0.062 g

  • mass of K₂CO₃ :

0.44 x 1.205 g = 0.5302 g

Mass C in K₂CO₃(MW=138 g/mol) :

= (12/138) x 0.5302

= 0.046 g

Total mass Of CO₂ :

= 0.062 + 0.046

= 0.108 g

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How many liters of carbon dioxide will be produced at STP if 3.56 g calcium carbonate reacts completely with carbon dioxide? CaC
sp2606 [1]

Answer:

V = 0.798 L

Explanation:

Hello there!

In this case, for this gas stoichiometry problem, we first need to compute the moles of carbon dioxide via stoichiometry and the molar mass of starting calcium carbonate:

3.56gCaCO_3*\frac{1molCaCO_3}{100gCaCO_3} *\frac{1molCO_2}{1molCaCO_3} =0.0356molCO_2

Next, we use the ideal gas equation for computing the volume, by bearing to mind that the STP conditions stand for a pressure of 1 atm and a temperature of 273.15 K:

PV=nRT\\\\V=\frac{nRT}{P}\\\\V=\frac{0.0356mol*0.08206\frac{atm*L}{mol*K}*273.15K}{1atm} \\\\V=0.798L

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4 0
3 years ago
Need help with this question.... Next to each Formula, write the number of atoms of each element found in one unit of the compou
Andrew [12]

Answer:

The number of atoms present in one unit of the following compounds is:

a). Potassium Iodide , KI = 2

b).Sodium Sulfide, Na_{2}S = 3

c). Silicon Dioxide , SiO_{2} = 3

d). Carbonic Acid ,H_{2}CO_{3} = 6

Explanation:

Atomicity : It is defined as the number of atoms that are present in a given molecule/compound.

Atom : The smallest unit of matter is called atom. For e.g O is atom of oxygen but O_{2} is not an atom , it is molecule of oxygen .

O_{2} molecule has 2 atoms of Oxygen

Similarly  Na , K , Fe are atoms but O_{2} ,CO_{2} ,N_{2},H_{2}  are molecules

a).Potassium Iodide

KI = 1 atom of K + 1 atom of I

Total atoms = 2

b) Sodium Sulfide

Na_{2}S = 2 atoms of Na + 1 atom of S

Total atoms = 3

c) Silicon Dioxide

SiO_{2} = 1 atom of Si + 2 atoms of O

Total atoms = 3

d) Carbonic Acid

H_{2}CO_{3} = 2 atom of H + 1 atom of C + 3 atom of O

= 2+1+3

Total atoms = 6

7 0
3 years ago
When maganese dioxide is added to hydrogen chloride you get water maganese dichloride and chlorine gas write the balanved equati
babunello [35]

Answer: When maganese dioxide is added to hydrogen chloride you get water maganese dichloride and chlorine gas then balanced equation is MnO_{2} + 4HCl \rightarrow 2H_{2}O + MnCl_{2} + Cl_{2}.

Explanation:

The word equation is given as maganese dioxide is added to hydrogen chloride you get water maganese dichloride and chlorine gas.

Now, in terms of chemical formulae this reaction equation will be as follows.

MnO_{2} + HCl \rightarrow H_{2}O + MnCl_{2} + Cl_{2}

Here, number of atoms on reactant side are as follows.

  • Mn = 1
  • O = 2
  • H = 1
  • Cl = 1

Number of atoms on product side are as follows.

  • Mn = 1
  • O = 1
  • H = 2
  • Cl = 4

To balance this equation, multiply HCl by 4 on reactant side and multiply H_{2}O by 2 on product side. Therefore, the equation can be rewritten as follows.

MnO_{2} + 4HCl \rightarrow 2H_{2}O + MnCl_{2} + Cl_{2}

Hence, number of atoms on reactant side are as follows.

  • Mn = 1
  • O = 2
  • H = 4
  • Cl = 4

Number of atoms on product side are as follows.

  • Mn = 1
  • O = 2
  • H = 4
  • Cl = 4

Since, this equation contains same number of atoms on both reactant and product side. Therefore, this equation is now balanced equation.

Thus, we can conclude that when maganese dioxide is added to hydrogen chloride you get water maganese dichloride and chlorine gas then balanced equation is MnO_{2} + 4HCl \rightarrow 2H_{2}O + MnCl_{2} + Cl_{2}.

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