16. how many grams of calcium will react with 44.8 liters of oxygen gas at standard conditions? 2 ca + o₂ →…

Explanation:

Step1: Calculate moles of oxygen gas

At standard - conditions, 1 mole of any gas occupies 22.4 L. So, the number of moles of $O_2$, $n_{O_2}=\frac{V}{V_m}$, where $V = 44.8$ L and $V_m=22.4$ L/mol. $n_{O_2}=\frac{44.8}{22.4}=2$ mol

Step2: Determine mole - ratio of calcium to oxygen

From the balanced chemical equation $2Ca + O_2\rightarrow2CaO$, the mole - ratio of $Ca$ to $O_2$ is $n_{Ca}:n_{O_2}=2:1$. So, if $n_{O_2} = 2$ mol, then $n_{Ca}=2\times n_{O_2}=4$ mol.

Step3: Calculate mass of calcium

The molar mass of calcium, $M_{Ca}=40$ g/mol. The mass of calcium, $m_{Ca}=n_{Ca}\times M_{Ca}$. $m_{Ca}=4\times40 = 160$ g

Answer:

160 g