Question

name ______________ quiz 1.1 intro to stoich solve the following problems. show your work or solution. observe correct number of significant figure(s). encircle or box your final answer. 1) a student has a 1 g sample of each of the following compounds: bacl₂, babr₂, kbr, kf. arrange the following in order of increasing number of moles in the sample. 2) convert 15.0 grams co to molecules? co to molecules = 15.0 grams co x 3) 15.0 l co₂ to moles to co₂ to molecules? co₂ molecules = 15.0 l co₂ x 4) how many moles of bacl₂ are there in 3.00 x 10²⁴ formula units bacl₂? (avogadro’s number = 6.022 x 10²³)

Explanation:

Step1: Calculate molar masses

Molar mass of \(BaCl_2\): \(M_{BaCl_2}=137.33 + 2\times35.45=208.23\ g/mol\)
Molar mass of \(BaBr_2\): \(M_{BaBr_2}=137.33+2\times79.90 = 297.13\ g/mol\)
Molar mass of \(KBr\): \(M_{KBr}=39.098 + 79.90=118.998\ g/mol\)
Molar mass of \(KF\): \(M_{KF}=39.098+18.998 = 58.096\ g/mol\)

Step2: Calculate moles for 1 - g samples

For \(BaCl_2\), \(n_{BaCl_2}=\frac{1\ g}{208.23\ g/mol}\approx0.00480\ mol\)
For \(BaBr_2\), \(n_{BaBr_2}=\frac{1\ g}{297.13\ g/mol}\approx0.00337\ mol\)
For \(KBr\), \(n_{KBr}=\frac{1\ g}{118.998\ g/mol}\approx0.00840\ mol\)
For \(KF\), \(n_{KF}=\frac{1\ g}{58.096\ g/mol}\approx0.0172\ mol\)
Increasing - order of moles: \(BaBr_2

Step3: Convert 15.0 g CO to moles

Molar mass of \(CO\) is \(M_{CO}=12.01 + 16.00=28.01\ g/mol\)
\(n_{CO}=\frac{15.0\ g}{28.01\ g/mol}\approx0.5355\ mol\)

Step4: Convert moles of CO to molecules

Number of molecules \(N = n\times N_A\), where \(N_A = 6.022\times10^{23}\ mol^{-1}\)
\(N=0.5355\ mol\times6.022\times10^{23}\ mol^{-1}\approx3.225\times10^{23}\) molecules

Step5: Convert 15.0 L \(CO_2\) to moles at STP

At STP (\(T = 273\ K\), \(P = 1\ atm\)), \(1\ mol\) of any gas occupies \(22.4\ L\)
\(n_{CO_2}=\frac{15.0\ L}{22.4\ L/mol}\approx0.670\ mol\)

Step6: Convert moles of \(CO_2\) to molecules

\(N=n_{CO_2}\times N_A=0.670\ mol\times6.022\times10^{23}\ mol^{-1}\approx4.035\times10^{23}\) molecules

Step7: Convert formula - units of \(BaCl_2\) to moles

\(n=\frac{\text{Number of formula - units}}{N_A}\)
\(n=\frac{3.00\times10^{24}}{6.022\times10^{23}\ mol^{-1}}\approx4.98\ mol\)

Answer:

1. \(BaBr_2
2. \(3.23\times 10^{23}\) molecules
3. \(4.04\times 10^{23}\) molecules
4. \(4.98\ mol\)