problem 4: percent yield\nquestion:\nin a reaction, 10.0 grams of caco₃ decomposes to produce cao and co₂…

Problem 4: Percent Yield

Explanation:

Step1: Calculate the molar mass of CaCO₃

The molar mass of CaCO₃: $M_{CaCO_3}=40.08 + 12.01+3\times16.00=100.09\ g/mol$

Step2: Calculate the moles of CaCO₃

$n_{CaCO_3}=\frac{m_{CaCO_3}}{M_{CaCO_3}}=\frac{10.0\ g}{100.09\ g/mol}\approx0.1\ mol$

Step3: Determine the theoretical yield of CaO

From the balanced - equation $CaCO_3\rightarrow CaO + CO_2$, the mole ratio of $CaCO_3$ to $CaO$ is 1:1. So the moles of CaO produced theoretically, $n_{CaO - theoretical}=n_{CaCO_3}=0.1\ mol$. The molar mass of CaO is $M_{CaO}=40.08 + 16.00 = 56.08\ g/mol$. Then the theoretical yield of CaO, $m_{CaO - theoretical}=n_{CaO - theoretical}\times M_{CaO}=0.1\ mol\times56.08\ g/mol = 5.608\ g$

Step4: Calculate the percent yield

The percent - yield formula is $\text{Percent Yield}=\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times100%$. Given actual yield $m_{CaO - actual}=4.0\ g$. So $\text{Percent Yield}=\frac{4.0\ g}{5.608\ g}\times100%\approx71.3%$

Answer:

$71.3%$

Problem 5: Volume of Gas at STP

Explanation:

Step1: Calculate the moles of H₂

The molar mass of H₂ is $M_{H_2}=2\times1.01\ g/mol = 2.02\ g/mol$. The moles of H₂, $n_{H_2}=\frac{m_{H_2}}{M_{H_2}}=\frac{5.0\ g}{2.02\ g/mol}\approx2.48\ mol$

Step2: Determine the moles of O₂ required

From the balanced - equation $2H_2(g)+O_2(g)\rightarrow2H_2O(l)$, the mole ratio of $H_2$ to $O_2$ is 2:1. So the moles of O₂ required, $n_{O_2}=\frac{1}{2}n_{H_2}=\frac{1}{2}\times2.48\ mol = 1.24\ mol$

Step3: Calculate the volume of O₂ at STP

At STP (Standard Temperature and Pressure, $T = 273\ K$, $P = 1\ atm$), the molar volume of a gas is $V_m = 22.4\ L/mol$. The volume of O₂, $V_{O_2}=n_{O_2}\times V_m=1.24\ mol\times22.4\ L/mol = 27.776\ L\approx27.8\ L$

Answer:

$27.8\ L$