a) Calculate the percentage of water in CaSO_(4)2H_(2)O The percentage composition of compound is 26.5% potassium (K) chromium (Cr) and 38.1% oxygon (O) Determine the empirical form compound. 6. A compound contains 0.70 gram of copper, 0.32 gram of sulfur, and oxygen. Calculate the empirical formula. 7. Propane, C_(3)H_(5) bums in oxygen according to the following equation C_(3)H_(1)+5O_(2)arrow 3CO_(2)+4H_(2)O A) How many moles of CO_(2) are produced when 25 moles of O_(2) rea B) How many moles of C_(3)H_(n) are needed to produce 5.76 moles of 8. What is the oxidation number of P in Mg_(3)(PO_(4))_(2)

Let's go through each question one by one:<br /><br />1. Calculate the percentage of water in $CaSO_{4}2H_{2}O$.<br /><br />To calculate the percentage of water in the compound, we need to find the molar mass of the compound and the molar mass of water, then divide the molar mass of water by the molar mass of the compound and multiply by 100.<br /><br />The molar mass of $CaSO_{4}2H_{2}O$ is:<br />$40.08 + 32.07 + 4(16.00) + 2(1.01 + 16.00) = 172.16 \text{ g/mol}$<br /><br />The molar mass of water is:<br />$2(1.01 + 16.00) = 18.02 \text{ g/mol}$<br /><br />The percentage of water in $CaSO_{4}2H_{2}O$ is:<br />$\frac{18.02}{172.16} \times 100 = 10.45\%$<br /><br />2. The percentage composition of a compound is $26.5\%$ potassium $(K)$, $38.1\%$ chromium $(Cr)$, and $35.4\%$ oxygen $(O)$. Determine the empirical formula of the compound.<br /><br />To determine the empirical formula, we need to find the moles of each element in a 100 g sample and then find the simplest whole number ratio.<br /><br />Moles of $K$:<br />$\frac{26.5}{39.10} = 0.676$<br /><br />Moles of $Cr$:<br />$\frac{38.1}{52.00} = 0.732$<br /><br />Moles of $O$:<br />$\frac{35.4}{16.00} = 2.213$<br /><br />Divide each by the smallest number of moles:<br />$\frac{0.676}{0.676} = 1$<br /><br />$\frac{0.732}{0.676} = 1.09$<br /><br />$\frac{2.213}{0.676} = 3.29$<br /><br />The empirical formula is $KCrO_{3}$.<br /><br />3. A compound contains 0.70 gram of copper, 0.32 gram of sulfur, and oxygen. Calculate the empirical formula.<br /><br />To calculate the empirical formula, we need to find the moles of each element and then find the simplest whole number ratio.<br /><br />Moles of $Cu$:<br />$\frac{0.70}{63.55} = 0.0111$<br /><br />Moles of $S$:<br />$\frac{0.32}{32.07} = 0.00999$<br /><br />Moles of $O$:<br />$\frac{0.01}{16.00} = 0.000625$<br /><br />Divide each by the smallest number of moles:<br />$\frac{0.0111}{0.000625} = 17.76$<br /><br />$\frac{0.00999}{0.000625} = 15.99$<br /><br />$\frac{0.000625}{0.000625} = 1$<br /><br />The empirical formula is $Cu_{17}S_{16}O$.<br /><br />4. Propane, $C_{3}H_{5}$, burns in oxygen according to the following equation:<br />$C_{3}H_{5} + 5O_{2} \rightarrow 3CO_{2} + 4H_{2}O$<br /><br />A) How many moles of $CO_{2}$ are produced when 25 moles of $O_{2}$ react?<br /><br />From the balanced equation, we can see that 5 moles of $O_{2}$ produce 3 moles of $CO_{2}$. Therefore, 25 moles of $O_{2}$ will produce:<br />$\frac{25}{5} \times 3 = 15$ moles of $CO_{2}$.<br /><br />B) How many moles of $C_{3}H_{5}$ are needed to produce 5.76 moles of $CO_{2}$?<br /><br />From the balanced equation, we can see that 3 moles of $CO_{2}$ are produced from 1 mole of $C_{3}H_{5}$. Therefore, to produce 5.76 moles of $CO_{2}$, we need:<br />$\frac{5.76}{3} = 1.92$ moles of $C_{3}H_{5}$.<br /><br />5. What is the oxidation number of P in $Mg_{3}(PO_{4})_{2}$?<br /><br />The oxidation number of magnesium (Mg) is +2. The oxidation number of oxygen (O) is -2. The compound is neutral, so the sum of the oxidation numbers of all the atoms in the compound is 0.<br /><br />Let the oxidation number of phosphorus (P) be x.<br /><br />$3(2) + 2