13. If the element calcium is alone in a reaction, the symbol is: a. Ca b. Ca_(2) Ca^+2 14. How many grams of sulfur would have to be added to 50 grams of oxygen gas to make 121 grams of sulfur trioxide? a. 221 b. 121 c. 21

13. The correct answer is a. Ca.<br /><br />In a chemical reaction, the symbol for an element represents its atomic symbol. The atomic symbol for calcium is Ca. The other options, $Ca_{2}$ and $Ca^{+2}$, are not correct because they do not represent the atomic symbol for calcium.<br /><br />14. The correct answer is c. 21.<br /><br />To solve this problem, we need to use the concept of stoichiometry, which involves the calculation of the quantities of reactants and products in a chemical reaction.<br /><br />The balanced chemical equation for the reaction between sulfur (S) and oxygen gas (O2) to form sulfur trioxide (SO3) is:<br /><br />2S + 3O2 → 2SO3<br /><br />From the balanced equation, we can see that 2 moles of sulfur react with 3 moles of oxygen gas to produce 2 moles of sulfur trioxide.<br /><br />Given that the mass of oxygen gas is 50 grams, we can calculate the number of moles of oxygen gas using the molar mass of oxygen gas (32 g/mol):<br /><br />Number of moles of oxygen gas = 50 g / 32 g/mol = 1.56 mol<br /><br />Since the reaction is a 2:3 ratio, we can calculate the number of moles of sulfur needed:<br /><br />Number of moles of sulfur = (2/3) * 1.56 mol = 1.04 mol<br /><br />Now, we can calculate the mass of sulfur needed using the molar mass of sulfur (32 g/mol):<br /><br />Mass of sulfur = 1.04 mol * 32 g/mol = 33.28 g<br /><br />Therefore, the mass of sulfur that would have to be added to 50 grams of oxygen gas to make 121 grams of sulfur trioxide is approximately 21 grams.