Question

How many molecules of acetylene(C₂H₂) would be formed when 10g of calcium carbide(CaC₂) is reacted with water(H₂O)? (Solve with Solution)
a) 9.4 x 10²²
b) 94 x 10²°
c) 9.4 x 10²³
d) 18.8 x 10²²

Answer 1

for this reaction is :-
CaC₂ + 2H₂O → Ca(OH)₂ + C₂H₂
two mole of water is react with 1 mole calcium carbide to form                                 1 mole of acetylene.
for 10 gm calcium carbide mole of acetylene is 
  =10 g CaC₂       = 0.1560 mol C₂H₂
    64.10 g CaC₂ [ M.W]   

1 mole of C₂H₂ contain 6.022×10²³[Avogadro's number] molecules
for 0.156 mol of C₂H₂ contain  = 0.156 x 6.022×10²³
                                            = 0.939 x 10²³
                                           ≈  9.4 x 10²² 

Answer 2

Hello!

How many molecules of acetylene(C₂H₂) would be formed when 10g of calcium carbide(CaC₂) is reacted with water(H₂O)? (Solve with Solution)

a) 9.4 x 10²²

b) 94 x 10²°

c) 9.4 x 10²³

d) 18.8 x 10²²

We have the following data:

MW(Molar Weight - CaC2) = 40u + 2*(12u) = 40u + 24u = 64u (64 g/mol)

We have the following double reaction balanced equation:

$CaC_2 + 2\:H_2O \Longrightarrow Ca(OH)_2 + C_2H_2$

If, 1 mol of CaC2 (64g) reacts with 2 moles of H2O forming 1 mol of acetylene (C2H2), then:

1 mol C2H2 ------------- 64 g CaC2

y mol C2H2 ------------- 10 g CaC2

$\dfrac{1}{y} = \dfrac{64}{10}$

multiply the means by the extremes

$64*y = 1*10$

$64\:y = 10$

$y = \dfrac{10}{64}$

$\boxed{y = 0.15625\:mol\:C_2H_2}$

Knowing that by Avogadro's Law for each mole of a substance we have 6.022 * 10²³ molecules. Then:

1 mol C2H2------------------ 6.022*10²³ molecules C2H2

0.15625 mol C2H2  -------------------- y molecules C2H2

$\dfrac{1}{0.15625} = \dfrac{6.022*10^{23}}{y}$

multiply the means by the extremes

$1*y = 0.15625*6.022*10^{23}$

$\boxed{\boxed{y \approx 9.4*10^{22}\:molecules\:of\:C_2H_2}}\:\:\:\:\:\:\bf\purple{\checkmark}$

Answer:

a) 9.4 x 10²²

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