Question #1e358

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Question #1e358

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Answer 1 · 2017-01-15T06:02:20

${CaC}_{2} + {2H}_{2} O \to C_{2} H_{2} + {Ca(OH)}_{2}$ $"CaC"_2 + "2H"_2"O" → "C"_2"H"_2 + "Ca(OH)"_2$

Explanation:

You follow a systematic procedure to balance the equation.

The other product is calcium hydroxide.

Start by converting the words to a chemical equation.:

${CaC}_{2} + H_{2} O \to C_{2} H_{2} + {Ca(OH)}_{2}$ $"CaC"_2 + "H"_2"O" → "C"_2"H"_2 + "Ca(OH)"_2$

A method that often works is first to balance everything other than $O$ $"O"$ and $H$ $"H"$ , then balance $O$ $"O"$ , and finally balance $H$ $"H"$ .

Another useful procedure is to start with what looks like the most complicated formula.

The most complicated formula looks like ${Ca(OH)}_{2}$ $"Ca(OH)"_2$ . We put a 1 in front of it to remind ourselves that the number is now fixed.

${CaC}_{2} + H_{2} O \to C_{2} H_{2} + 1 {Ca(OH)}_{2}$ $"CaC"_2 + "H"_2"O" → "C"_2"H"_2 + color(red)(1)"Ca(OH)"_2$

Balance $Ca$ $"Ca"$ :

We have $1 Ca$ $"1 Ca"$ on the right, so we need $1 Ca$ $"1 Ca"$ on the left. We put a 1 in front of the ${CaC}_{2}$ $"CaC"_2$ .

$1 {CaC}_{2} + H_{2} O \to C_{2} H_{2} + 1 {Ca(OH)}_{2}$ $color(orange)(1)"CaC"_2 + "H"_2"O" → "C"_2"H"_2 + color(red)(1)"Ca(OH)"_2$

Balance $C$ $"C"$ :

We have fixed $2 C$ $"2 C"$ on the left. We need $2 C$ $"2 C"$ on the right. Put a 1 in front of $C_{2} H_{2}$ $"C"_2"H"_2$ .

$1 {CaC}_{2} + H_{2} O \to 1 C_{2} H_{2} + 1 {Ca(OH)}_{2}$ $color(orange)(1)"CaC"_2 + "H"_2"O" → color(blue)(1)"C"_2"H"_2 + color(red)(1)"Ca(OH)"_2$

Balance $O$ $"O"$ :

We have fixed $2 O$ $"2 O"$ on the right. We need $2 O$ $"2 O"$ on the left. Put a 2 in front of $H_{2} O$ $"H"_2"O"$ .

$1 {CaC}_{2} + 2 H_{2} O \to 1 C_{2} H_{2} + 1 {Ca(OH)}_{2}$ $color(orange)(1)"CaC"_2 + color(purple)(2)"H"_2"O" → color(blue)(1)"C"_2"H"_2 + color(red)(1)"Ca(OH)"_2$

Every formula now has a coefficient. We should have a balanced equation.

Let's check.

$Atom m lhs m rhs$ $bb("Atom" color(white)(m)"lhs"color(white)(m)"rhs")$
$m Ca m m l 1 m m l 1$ $color(white)(m)"Ca"color(white)(mml)1color(white)(mml)1$
$m C m m m 2 m m l 2$ $color(white)(m)"C"color(white)(mmm)2color(white)(mml)2$
$m H m m m 4 m m l 4$ $color(white)(m)"H"color(white)(mmm)4color(white)(mml)4$
$m O m m m 2 m m l 2$ $color(white)(m)"O"color(white)(mmm)2color(white)(mml)2$

All atoms balance. The balanced equation is

${CaC}_{2} + {2H}_{2} O \to C_{2} H_{2} + {Ca(OH)}_{2}$ $"CaC"_2 + "2H"_2"O" → "C"_2"H"_2 + "Ca(OH)"_2$

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Question #1e358

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