Question

if 15.0g of caco₃ is heated, the theoretical yield of cao is 8.40g. if the actual yield is 6.10g, what is the percent yield? your answer what is the mass of 0.250 mol of sodium bromide? your answer how many grams of co₂ are produced if15.0 ml of ethanol fully combust? ethanols density is 0.789 g/ml. your answer how many moles are in 94.0g of carbon dioxide?

Explanation:

Step1: Calculate the percent yield formula

Percent yield = $\frac{\text{actual yield}}{\text{theoretical yield}}\times100\%$

Step2: Substitute the values

Given actual yield = $6.10$ g, theoretical yield = $8.40$ g.
Percent yield = $\frac{6.10}{8.40}\times100\%$

Step3: Calculate the value

$\frac{6.10}{8.40}\times100\% \approx 72.6\%$

Step1: Find the molar mass of sodium bromide ($NaBr$)

Molar mass of $Na$ = $22.99$ g/mol, molar mass of $Br$ = $79.90$ g/mol.
Molar mass of $NaBr$ = $22.99 + 79.90=102.89$ g/mol

Step2: Use the formula $m = n\times M$

where $n = 0.250$ mol (amount of substance), $M = 102.89$ g/mol (molar mass)
$m=0.250\times102.89 = 25.7$ g

Step1: Calculate the mass of ethanol

Density formula $
ho=\frac{m}{V}$, so $m=
ho\times V$.
Given $
ho = 0.789$ g/mL, $V = 15.0$ mL.
$m = 0.789\times15.0=11.835$ g

Step2: Find the molar mass of ethanol ($C_2H_5OH$)

Molar mass of $C$ = $12.01$ g/mol, $H$ = $1.01$ g/mol, $O$ = $16.00$ g/mol.
Molar mass of $C_2H_5OH$ = $2\times12.01+6\times1.01 + 16.00=46.08$ g/mol

Step3: Calculate the moles of ethanol

$n=\frac{m}{M}=\frac{11.835}{46.08}\approx0.257$ mol

Step4: Write the combustion reaction of ethanol

$C_2H_5OH+3O_2
ightarrow2CO_2 + 3H_2O$
From the reaction, 1 mole of ethanol produces 2 moles of $CO_2$.
So moles of $CO_2$ produced = $2\times0.257 = 0.514$ mol

Step5: Find the molar mass of $CO_2$

Molar mass of $CO_2$ = $12.01+2\times16.00 = 44.01$ g/mol

Step6: Calculate the mass of $CO_2$

$m=n\times M=0.514\times44.01\approx22.6$ g

Step1: Find the molar mass of $CO_2$

Molar mass of $CO_2$ = $12.01+2\times16.00 = 44.01$ g/mol

Step2: Use the formula $n=\frac{m}{M}$

Given $m = 94.0$ g, $M = 44.01$ g/mol
$n=\frac{94.0}{44.01}\approx2.14$ mol

Answer:

$72.6\%$