name ______________ quiz 1.1 intro to stoich solve the following problems. show your work or solution…

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)

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)

Answer

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<BaCl_2<KBr<KF)

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<BaCl_2<KBr<KF)
  2. (3.23\times 10^{23}) molecules
  3. (4.04\times 10^{23}) molecules
  4. (4.98\ mol)