16. If the equation is correctly balanced, what are the coefficients? \( \begin{array}{ll}\text { A. } 2: 3 ; 2: 5 & \text { C. } 1 ; 5 ; 3 ; 4 \\ \text { B. } 2 ; 5 ; 1: 4 & \text { D. } 3 ; 5 ; 4 ; 10\end{array} \) 17. If 50.0 g of \( \mathrm{C}_{3} \mathrm{H}_{8} \) is allowed to react with 200 g of \( \mathrm{O}_{2} \), which is the limiting reagent? (2points) \( \begin{array}{ll}\text { A. } \mathrm{C}_{3} \mathrm{H}_{8} & \text { C. } \mathrm{CO}_{2} \\ \text { B. } \mathrm{O}_{2} & \text { D. } \mathrm{H}_{2} \mathrm{O}\end{array} \) 18. How many moles of carbon dioxide can be produced from the reaction? (2points) \( \begin{array}{ll}\text { A. } 1.00 \mathrm{~mol} & \text { C. } 150 \mathrm{~mol} \\ \text { B. } 1.14 \mathrm{~mol} & \text { D. } 3.41 \mathrm{~mol} \\ \text { 19. How much of the excess reactant would be left unused? }(2 \text { points) } \\ \text { A. } 18.0 \mathrm{~g} & \text { C. } 5.00 \mathrm{~g}\end{array} \)
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Mind Expander
When balancing chemical equations, it’s always a good idea to start with the most complex molecule present. For combustion reactions like the one involving \( \mathrm{C}_3\mathrm{H}_8 \) (propane) and \( \mathrm{O}_2 \), remember to prioritize balancing carbon (C) and hydrogen (H) before adjusting oxygen (O). It can feel like a puzzle, but once you find the right coefficients, everything will click into place! To find the limiting reagent between \( \mathrm{C}_3\mathrm{H}_8 \) and \( \mathrm{O}_2 \), you need to convert mass into moles using molar mass. For propane, it’s about 44.1 g/mol, yielding approximately 1.14 moles from 50.0 g. For oxygen, 200 g converts to about 6.22 moles (32 g/mol). By comparing mole ratios in the balanced equation, you'll uncover which reactant runs out first – the essence of determining the limiting reactant!
