Start by balancing the given chemical equation:

\(\displaystyle \text{C}_5\text{H}_{12}(g)+8\text{O}_2(g)\rightarrow 5\text{CO}_2(g)+6\text{H}_2\text{O}(g)\)

Next, convert the grams of pentane into moles of pentane.

\(\displaystyle 152.3\text{g }\text{C}_5\text{H}_{12}\cdot\frac{\text{1 molC}_5\text{H}_{12}}{72.15\text{g C}_5\text{H}_{12}}=2.110 \text{ moles of C}_5\text{H}_{12}\)

Use the stoichiometric ratio given by the balanced equation to find the number of moles of oxygen needed to react with the pentane.

\(\displaystyle 2.110\text{ moles of C}_5\text{H}_{12}\cdot\frac{8\text{ moles of O}_2}{1\text{mole of C}_5\text{H}_{12}}=16.89\text{ moles of O}_2\)

Finally, convert the number of moles of oxygen into grams of oxygen.

\(\displaystyle 16.89\text{ moles of O}_2\cdot \frac{32.00\text{g of O}_2}{1\text{ mole of O}_2}=540.39\text{g of O}_2\)

There should be \(\displaystyle 4\) significant figures, so the answer is \(\displaystyle 540.4\text{g of O}_2\).

Start by balancing the equation:

\(\displaystyle \text{K}_2\text{SO}_4(aq)+\text{BaCl}_2(aq)\rightarrow 2\text{KCl}(aq)+\text{BaSO}_4(s)\)

Next, figure out which reactant is the limiting reactant.

\(\displaystyle 15.0\times10^{-3}\text{L K}_2\text{SO}_4\cdot\frac{1.20\text{ moles of K}_2\text{SO}_4}{1.00\text{L K}_2\text{SO}_4}\cdot\frac{\text{1 mole K}_2\text{SO}_4}{\text{1 mole BaSO}_4}=0.018\text{ moles of BaSO}_4\)

\(\displaystyle 25.0\times10^{-3}\text{L BaCl}_2\cdot\frac{0.75\text{moles BaCl}_2}{1\text{L BaCl}_2}\cdot\frac{1\text{ mole BaCl}_2}{1\text{ mole BaSO}_4}=0.01875\text{ moles of BaSO}_4\)

Since fewer moles of \(\displaystyle \text{BaSO}_4\) is produced when \(\displaystyle \text{K}_2\text{SO}_4\) is reacted, then \(\displaystyle \text{K}_2\text{SO}_4\) must be the limiting reactant.

Convert the amount of moles of \(\displaystyle \text{BaSO}_4\) into grams.

\(\displaystyle 0.018\text{ moles of BaSO}_4\cdot\frac{233.43\text{g BaSO}_4}{\text{1 mole BaSO}_4}=4.20\text{g BaSO}_4\)

For this question, we're given the density and volume of a gas and we're asked to find the mass of the gas.

To answer this question, we'll need to use dimensional analysis. What this means is that we'll need to cancel out units in order to obtain the ones that we're looking for.

\(\displaystyle 2.5\: L\cdot\frac{1000\:mL}{1\:L}\cdot \frac{0.76\: g}{1\: mL}\cdot \frac{1\:kg}{1000\:g}=1.9\:kg\)

Use the relations \(\displaystyle 1mol=22.4L H_{2}\), molar mass \(\displaystyle H_{2}O=18.02\frac{g}{mol}\), and \(\displaystyle 2molH_{2}O=2molH_{2}\)

\(\displaystyle 2.34L H_{2} * \frac{1mol H_{2}}{22.4L H_{2}}=.104mol H_{2} * \frac{2mol H_{2}O}{2molH_{2}}=.104molH_{2}O*\frac{18.02gH_{2}O}{1molH_{2}O}=1.88gH_{2}O\)