Question

How many molecule of acetylene gas c2h2 would be produced in the reaction of 10.0g of calcium carbide with excess of water?

Answer 1

Answer in the attachment
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Answer 2

Hello!

How many molecule of acetylene gas c2h2 would be produced in the reaction of 10.0 g of calcium carbide with excess of water?

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:

9.4 x 10²²

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