# Question 5f0a2

May 11, 2016

$\text{Grams of acetylene}$ $\cong$ $2 \cdot g$

#### Explanation:

You have done the hard yards when you have the stoichiometric equation, which I note you have done:

$C a C \equiv C \left(s\right) + 2 {H}_{2} O \left(l\right) \rightarrow H C \equiv C H + C a {\left(O H\right)}_{2}$

This equation explicitly says that $64.1 \cdot g$ of calcium carbide react with $36 \cdot g$ of water to produce $26 \cdot g$ acetylene and $74 \cdot g$ calcium hydroxide.

Each of these numbers represents molar equivalents of the reactants and products.

You started with $\frac{5.00 \cdot g}{64.1 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.0780 \cdot m o l$ of calcium carbide. If follows that I formed an equivalent quantity of acetylene, i.e. $0.0780 \cdot m o l$, and of calcium hydroxide.

So $\text{mass of acetylene = molar quantity"xx"molar mass}$ $=$ 0.0780*molxx26.0*g*mol^-1=??*g#

This reaction was the basis of the carbide lamp.