15. how many grams of calcium must react with of sulfuric acid, h₂so₄, in order to produce 5.6 liters of…

Explanation:

Step1: Determine moles of hydrogen gas

At standard - temperature and pressure (STP), 1 mole of any gas occupies 22.4 L. Given the volume of hydrogen gas $V = 5.6$ L. The number of moles of hydrogen gas $n_{H_2}$ is calculated using the formula $n=\frac{V}{V_m}$, where $V_m = 22.4$ L/mol. $n_{H_2}=\frac{5.6}{22.4}=0.25$ mol

Step2: Use mole - ratio from the balanced equation

The balanced chemical equation is $Ca + H_2SO_4\rightarrow CaSO_4 + H_2$. The mole - ratio of $Ca$ to $H_2$ is 1:1. So, the number of moles of calcium $n_{Ca}$ that reacts is equal to the number of moles of hydrogen gas produced. So, $n_{Ca}=n_{H_2}=0.25$ mol.

Step3: Calculate mass of calcium

The molar mass of calcium $M_{Ca}=40.08$ g/mol. The mass of calcium $m_{Ca}$ is calculated using the formula $m = n\times M$. $m_{Ca}=n_{Ca}\times M_{Ca}=0.25\times40.08 = 10.02$ g

Answer:

10.02 g