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Paha777 [63]
4 years ago
12

Calcium oxide and oxygen gas are produced by the thermal decomposition of limestone in the reaction CaCO (s) CaO(s) + CO (g). Wh

at mass of lime can be produced from 4.5 10 kg of limestone? Please show all the work
Chemistry
2 answers:
kozerog [31]4 years ago
6 0
The mass of lime that  can  be produced  from  4.510 Kg of limestone  is calculated  as  below

calculate the moles  of CaCO3  used

that is  moles =mass/molar mass
convert  Kg  to g  = 4.510 x1000 =4510g
=  4510 / 100 =45.10 moles
CaCO3 = CaO  +O2

by use of mole ratio between CaCO3  to CaO  (1:1) the  moles of CaO  is also= 45.10 moles
mass of CaO = moles x molar  mass

45.10  x56 = 2525.6  g  of  CaO



lutik1710 [3]4 years ago
6 0
As the power on mass of CaCO₃ is not given so we are assuming it 4.5 × 10³ Kg.

Answer:
             2520 Kg of CaO

Solution:
 
The balance chemical equation is as follow,

                                      CaCO₃    →    CaO  + CO₂

According to equation,

           0.1 Kg (1 mole) CaCO₃ produces  =  0.056 Kg (1 mole)  of CaO
So,
            4.5 × 10³ Kg CaCO₃ will produce  =  X Kg of CaO

Solving for X,
                     X  =  (4.5 × 10³ Kg × 0.056 Kg) ÷ 0.1 Kg

                     X  =  2520 Kg of CaO
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If 5.85 grams of cobalt metal react with 15.8 grams of silver nitrate, how many grams of silver metal can be formed and how many
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Answers:
<span>Answer 1: 10.03 g of siver metal can be formed.</span>
Answer 2: 3.11 g of Co are left over.

Work:

1) Unbalanced chemical equation (given):

<span>Co + AgNO3 → Co(NO3)2 + Ag

2) Balanced chemical equation
</span>
<span>Co + 2AgNO3 → Co(NO3)2 + 2Ag

3) mole ratios

1 mol Co : 2 mole AgNO3 : 1 mol Co(NO3)2 : 2 mol Ag

4) Convert the masses in grams of the reactants into number of moles

4.1) 5.85 grams of Co

# moles = mass in grams / atomic mass

atomic mass of Co = 58.933 g/mol

# moles Co = 5.85 g / 58.933 g/mol = 0.0993 mol

4.2) 15.8 grams of Ag(NO3)

# moles Ag(NO3) = mass in grams / molar mass

molar mass AgNO3 = 169.87 g/mol

# moles Ag(NO3) = 15.8 g / 169.87 g/mol = 0.0930 mol

5) Limiting reactant

Given the mole ratio 1 mol Co : 2 mol Ag(NO3) you can conclude that there is not enough Ag(NO3) to make all the Co react.

That means that Ag(NO3) is the limiting reactant, which means that it will be consumed completely, whilce Co is the excess reactant.

6) Product formed.

Use this proportion:

2 mol Ag(NO3)           0.0930mol Ag(NO3)    
--------------------- =      ---------------------------
    2 mol Ag                              x

=> x = 0.0930 mol

Convert 0.0930 mol Ag to grams:

mass Ag = # moles * atomic mass = 0.0930 mol * 107.868 g/mol = 10.03 g

Answer 1: 10.03 g of siver metal can be formed.

6) Excess reactant left over

    1 mol Co                             x
----------------------- =  ----------------------------
2 mole Ag(NO3)       0.0930 mol Ag(NO3)

=> x = 0.0930 / 2 mol Co = 0.0465 mol Co reacted

Excess = 0.0993 mol - 0.0465 mol = 0.0528 mol

Convert to grams:

0.0528 mol * 58.933 g/mol = 3.11 g

Answer 2: 3.11 g of Co are left over.
</span>


8 0
3 years ago
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