Can you show how to get the grams of salt that are formed after the reaction is

Question

Can you show how to get the grams of salt that are formed after the reaction is complete? Use dimensional analysis if possible.

A sample of 8.83 g of solid calcium hydroxide is added to 22.0 mL of 0.490 M aqueous hydrochloric acid. Enter the balanced chemical equation for the reaction. Physical states are optional and not graded. Ca(OH)2+2HCl CaCl+2H2O Tip: If you need to clear your work and reset the equation, click the button that looks like two arrows. What is the limiting reactant? O calcium hydroxide O hydrochloric acid w many grams of salt are formed after the reaction is complete? Number 611 How many grams of the excess reactant remain after the reaction is complete? Number 844

Explanation / Answer

1)

Molar mass of Ca(OH)2 = 1*MM(Ca) + 2*MM(O) + 2*MM(H)

= 1*40.08 + 2*16.0 + 2*1.008

= 74.096 g/mol

mass of Ca(OH)2 = 8.83 g

we have below equation to be used:

number of mol of Ca(OH)2,

n = mass of Ca(OH)2/molar mass of Ca(OH)2

=(8.83 g)/(74.096 g/mol)

= 0.1192 mol

volume of HCl, V = 22.0 mL

= 2.2*10^-2 L

we have below equation to be used:

number of mol in HCl,

n = Molarity * Volume

= 0.49*0.022

= 1.078*10^-2 mol

we have the Balanced chemical equation as:

Ca(OH)2 + 2 HCl ---> CaCl2 + 2 H2O

1 mol of Ca(OH)2 reacts with 2 mol of HCl

for 0.1192 mol of Ca(OH)2, 0.2383 mol of HCl is required

But we have 1.078*10^-2 mol of HCl

so, HCl is limiting reagent

2)

we will use HCl in further calculation

Molar mass of CaCl2 = 1*MM(Ca) + 2*MM(Cl)

= 1*40.08 + 2*35.45

= 110.98 g/mol

From balanced chemical reaction, we see that

when 2 mol of HCl reacts, 1 mol of CaCl2 is formed

mol of CaCl2 formed = (1/2)* moles of HCl

= (1/2)*1.078*10^-2

= 5.39*10^-3 mol

we have below equation to be used:

mass of CaCl2 = number of mol * molar mass

= 5.39*10^-3*1.11*10^2

= 0.5982 g

Answer: 0.598 g

3)

From balanced chemical reaction, we see that

when 2 mol of HCl reacts, 1 mol of Ca(OH)2 reacts

mol of Ca(OH)2 reacted = (1/2)* moles of HCl

= (1/2)*1.078*10^-2

= 5.39*10^-3 mol

mol of Ca(OH)2 remaining = mol initially present - mol reacted

mol of Ca(OH)2 remaining = 0.1192 - 5.39*10^-3

mol of Ca(OH)2 remaining = 0.1138 mol

Molar mass of Ca(OH)2 = 1*MM(Ca) + 2*MM(O) + 2*MM(H)

= 1*40.08 + 2*16.0 + 2*1.008

= 74.096 g/mol

we have below equation to be used:

mass of Ca(OH)2,

m = number of mol * molar mass

= 0.1138 mol * 74.096 g/mol

= 8.43 g

Answer: 8.43 g

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