Question

how much c2h2 would form if 2.4 grams of c reacted with 0.2 gram of h2?

reaction	reactant(s)	product(s)
mass	2.4 g	0.2 g	?

Explanation:

Step1: Calculate moles of C

The molar - mass of C is approximately 12 g/mol. The number of moles of C, $n_{C}=\frac{m_{C}}{M_{C}}$, where $m_{C} = 2.4$ g and $M_{C}=12$ g/mol. So, $n_{C}=\frac{2.4}{12}=0.2$ mol.

Step2: Calculate moles of $H_2$

The molar - mass of $H_2$ is approximately 2 g/mol. The number of moles of $H_2$, $n_{H_2}=\frac{m_{H_2}}{M_{H_2}}$, where $m_{H_2}=0.2$ g and $M_{H_2} = 2$ g/mol. So, $n_{H_2}=\frac{0.2}{2}=0.1$ mol.

Step3: Determine the limiting reactant

The balanced chemical equation for the formation of $C_2H_2$ from C and $H_2$ is $2C + H_2
ightarrow C_2H_2$. From the equation, the mole ratio of C to $H_2$ is 2:1. For 0.2 mol of C, we need 0.1 mol of $H_2$ (since $\frac{0.2}{2}=0.1$), and we have 0.1 mol of $H_2$. So, neither reactant is in excess.

Step4: Calculate moles of $C_2H_2$ formed

From the balanced equation, the mole ratio of $H_2$ to $C_2H_2$ is 1:1 and the mole ratio of C to $C_2H_2$ is 2:1. Since the reaction goes to completion with the given amounts of reactants, the number of moles of $C_2H_2$ formed is 0.1 mol.

Step5: Calculate mass of $C_2H_2$ formed

The molar - mass of $C_2H_2$ is $M_{C_2H_2}=2\times12 + 2\times1=26$ g/mol. The mass of $C_2H_2$, $m_{C_2H_2}=n_{C_2H_2}\times M_{C_2H_2}$, where $n_{C_2H_2}=0.1$ mol and $M_{C_2H_2}=26$ g/mol. So, $m_{C_2H_2}=0.1\times26 = 2.6$ g.

Answer:

2.6 g