Question

1. how long will it take (in seconds) for 45.0g of mg to react with hcl, if the average rate of the reaction is 2.30g mg/min? why is this not realistic? 5) electrolysis of h2o produces o2 and h2. if o2 gas is produced at an average rate of 28.5ml/min, calculate the consumption of h2o in grams per second. 6) given the reaction: h2(g) + cl2(g) ⇒ 2hcl(g) a) if 2.32g of hcl are produced in 4.0 min, what is the rate of reaction in mol hcl/second? b) if h2 is used up at a rate of 30.0 mol/s, at what rate is hcl produced in g/min? 7) c5h12(g) + 8o2(g) ⇒ 5co2(g) + 6h2o(g) if 17.6g of c5h12 is burned in three seconds, calculate the rate of the reaction in grams of co2 per second.

Explanation:

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Step1: Convert the rate to g Mg/s

The rate is 2.30 g Mg/min. To convert to g Mg/s, divide by 60. So the rate $r=\frac{2.30}{60}\text{ g/s}\approx0.0383\text{ g/s}$.

Step2: Calculate the time

We know that rate $r = \frac{\Delta m}{\Delta t}$, so $\Delta t=\frac{\Delta m}{r}$. Given $\Delta m = 45.0\text{ g}$ and $r = 0.0383\text{ g/s}$, then $\Delta t=\frac{45.0}{0.0383}\text{ s}\approx1175\text{ s}$.
This is not realistic because in real - life, the reaction rate of magnesium with hydrochloric acid is much faster due to the high reactivity of magnesium and the exothermic nature of the reaction which can increase the reaction rate over time.

Step1: Write the electrolysis equation of water

$2H_2O(l)\xrightarrow{electrolysis}2H_2(g)+O_2(g)$. The molar ratio of $H_2O$ to $O_2$ is 2:1.

Step2: Convert the rate of $O_2$ production to L/s

The rate of $O_2$ production is $v = 28.5\text{ mL/min}=\frac{28.5}{1000\times60}\text{ L/s}=4.75\times 10^{-4}\text{ L/s}$.
Assume $O_2$ is an ideal gas, at standard conditions ($T = 273\text{ K},P = 1\text{ atm}$), $n=\frac{PV}{RT}$. At STP, $P = 1\text{ atm},T=273\text{ K},R = 0.0821\text{ L}\cdot\text{atm}/(\text{mol}\cdot\text{K})$. For $O_2$, $n_{O_2}=\frac{1\times4.75\times 10^{-4}}{0.0821\times273}\text{ mol/s}\approx2.12\times 10^{-5}\text{ mol/s}$.

Step3: Calculate the rate of $H_2O$ consumption

Since the molar ratio of $H_2O$ to $O_2$ is 2:1, the rate of $H_2O$ consumption $n_{H_2O}=2\times n_{O_2}=4.24\times 10^{-5}\text{ mol/s}$.
The molar mass of $H_2O$ is $M = 18\text{ g/mol}$, so the mass - consumption rate of $H_2O$ is $m = n_{H_2O}\times M=4.24\times 10^{-5}\text{ mol/s}\times18\text{ g/mol}=7.63\times 10^{-4}\text{ g/s}$.

Step1: Calculate the number of moles of $HCl$ produced

The molar mass of $HCl$ is $M_{HCl}=1.01 + 35.45=36.46\text{ g/mol}$. Given $m_{HCl}=2.32\text{ g}$, then $n_{HCl}=\frac{m_{HCl}}{M_{HCl}}=\frac{2.32}{36.46}\text{ mol}\approx0.0636\text{ mol}$.

Step2: Convert the time to seconds

The time $t = 4.0\text{ min}=4\times60\text{ s}=240\text{ s}$.

Step3: Calculate the rate of reaction

The rate of reaction $r=\frac{n_{HCl}}{t}=\frac{0.0636}{240}\text{ mol/s}\approx2.65\times 10^{-4}\text{ mol/s}$.

Answer:

1175 s

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