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zepelin [54]
3 years ago
5

How many grams of H2O are produced when 35.0 g of NaOH reacts with 17.5 g of CO,?

Chemistry
1 answer:
zhenek [66]3 years ago
6 0

Answer:

2NaOH + CO2 -> Na2CO3 + H2O

1) Find the moles of each substance

\eq n(NaOH)=\frac{35.0}{22.99+16.00+1.008\\}\  =\frac{35.0}{39.998} \ = 0.8750437522 moles\\n(CO_{2} ) = \frac{17.5}{12.01+32.00} = \frac{17.5}{44.01} = 0.3976369007 moles\\

2) Determine the limitting reagent

\\NaOH = \frac{0.8750437522}{2} = 0.4375218761\\\\

∴ Carbon dioxide is limitting as it has a smaller value.

3) multiply the limiting reagent by the mole ratio of unknown over known

n(H2O ) = 0.3976369007 × 1/2

             = 0.1988184504 moles

4) Multiply the number of moles by the molar mass of the substance.

m = 0.1988184504 × (1.008 × 2 + 16.00)

   = 0.1988184504 × 18.016

   = 3.581913202 g

Explanation:

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HACTEHA [7]

Answer:

14.33 g

Explanation:

Solve this problem based on the stoichiometry of the reaction.

To do that we need the molecular weight of the masses involved and then calculate the number of moles, find the limiting reagent and  finally calculate the mass of AgCl.

                      2 AgNO₃   + CaCl₂    ⇒   Ca(NO₃)₂   + 2 AgCl

mass, g                  6.97        6.39                                     ?

MW ,g/mol         169.87      110.98                                  143.32

mol =m/MW           0.10         0.06                                    0.10

From the table above AgNO₃ is the limiting reagent and we will produce 0.10 mol AgCl which is a mass :

0.10 mol x 143.32 g/mol = 14.33 g

5 0
4 years ago
Name the type of equation shown here: 2NaBr+Ca(OH)2 —>CaBr2+2NaOH
sergeinik [125]
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6 0
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tresset_1 [31]

Answer:

The molality of isoborneol in camphor is 0.53 mol/kg.

Explanation:

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Melting point of sample = T_f = 165°C

Depression in freezing point = \Delta T_f=?

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Depression in freezing point  is also given by formula:

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14^oC=1\times 40^oC kg/mol\times m

m=\frac{14^oC}{1\times 40^oC kg/mol}=0.35 mol/kg

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8 0
3 years ago
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3 0
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