Question

question 19 using the equation 2li + 2h₂o → 2lioh + h₂ when 2.5 g of li reacts with water 0.34 g of h₂ is produced. the percent yield of the reaction is blank 1% blank 1 add your answer 5 points question 20 when your body burns 1.00 g of sucrose, c₁₂h₂₂o₁₁ + 12 o₂ → 12co₂ + 11 h₂o. blank 1 grams of oxygen is burned, according to the following reaction. blank 1 add your answer 5 points

Explanation:

Question 19:

Step1: Calculate moles of Li

First, find the molar - mass of Li. The molar mass of Li is approximately $M_{Li}=6.94\ g/mol$. The number of moles of Li, $n_{Li}=\frac{m_{Li}}{M_{Li}}$, where $m_{Li} = 2.5\ g$. So, $n_{Li}=\frac{2.5\ g}{6.94\ g/mol}\approx0.360\ mol$.

Step2: Determine theoretical moles of $H_2$

From the balanced chemical equation $2Li + 2H_2O
ightarrow2LiOH + H_2$, the mole - ratio of $Li$ to $H_2$ is 2:1. So, the theoretical number of moles of $H_2$, $n_{H_2,theo}=\frac{1}{2}n_{Li}=\frac{1}{2}\times0.360\ mol = 0.180\ mol$.

Step3: Calculate theoretical mass of $H_2$

The molar mass of $H_2$ is $M_{H_2}=2.02\ g/mol$. The theoretical mass of $H_2$, $m_{H_2,theo}=n_{H_2,theo}\times M_{H_2}=0.180\ mol\times2.02\ g/mol = 0.364\ g$.

Step4: Calculate percent yield

The percent yield, $\%yield=\frac{m_{H_2,actual}}{m_{H_2,theo}}\times100\%$, where $m_{H_2,actual}=0.34\ g$. So, $\%yield=\frac{0.34\ g}{0.364\ g}\times100\%\approx93.4\%$.

Question 20:

Step1: Calculate moles of sucrose

The molar mass of sucrose ($C_{12}H_{22}O_{11}$) is $M = 12\times12.01+22\times1.01 + 11\times16.00=342.3\ g/mol$. The number of moles of sucrose, $n=\frac{m}{M}$, where $m = 1.00\ g$. So, $n=\frac{1.00\ g}{342.3\ g/mol}\approx0.00292\ mol$.

Step2: Determine moles of $O_2$

From the balanced chemical equation $C_{12}H_{22}O_{11}+12O_2
ightarrow12CO_2 + 11H_2O$, the mole - ratio of $C_{12}H_{22}O_{11}$ to $O_2$ is 1:12. So, the number of moles of $O_2$, $n_{O_2}=12\times n = 12\times0.00292\ mol=0.0350\ mol$.

Step3: Calculate mass of $O_2$

The molar mass of $O_2$ is $M_{O_2}=32.00\ g/mol$. The mass of $O_2$, $m_{O_2}=n_{O_2}\times M_{O_2}=0.0350\ mol\times32.00\ g/mol = 1.12\ g$.

Answer:

Question 19: 93.4%
Question 20: 1.12