Question

1) Calculate the mass of calcium oxide residue gotten when 82g of calcium carbonate is heated to a steady mass.

2)Calculate the volume hydrogen produced when 6g of magnesium reacts with excess dilute hydrochloric acid at s.t.p

Answer 1

Calculate the mass of calcium oxide residue gotten when 82g of calcium carbonate is heated to a steady mass.

Calcium carbonate can be decomposed in a next way:

$$\mathrm{CaCO_3} \rightarrow \mathrm{CaO} + \mathrm{CO_2}$$

As can you see, the mole ratio between calcium carbonate and calcium oxide is 1: 1, it means that one mole of calcium carbonate can produce one mole of calcium oxide.

n = m/Mw, where n is amount, m is mass and Mw is molecular weight

$$\mathrm{n_1} = \mathrm{n_2}$$ $$\mathrm{m_1/Mw_1} = \mathrm{m_2/Mw_2}$$ $$\mathrm{m_1} = 82 \mathrm{g}$$ $$\mathrm{Mw_1} = 100 \mathrm{g/mol}$$ $$\mathrm{m_2} = \mathrm{X}$$ $$\mathrm{Mw_2} = 56$$ $$\mathrm{X} = (82/100) * 56 = 45,92 \mathrm{g}$$

Calculate the volume hydrogen produced when 6g of magnesium reacts with excess dilute hydrochloric acid at s.t.p

One mole of any gas at stp contains volume of 22.4 L

Let's find volume of hydrogen that produced when 6g of magnesium reacts with excess dilute hydrochloric acid at s.t.p

$$\mathrm{Mg} + 2\mathrm{HCl} = \mathrm{MgCl_2} + \mathrm{H_2}$$

As can you see, the mole ratio between magnesium and hydrogen is 1: 1, it means that one mole of magnesium can produce one mole of hydrogen.

$$\mathrm{n_1} = \mathrm{n_2}$$ $$\mathrm {m} _ {1} / \mathrm {M w} _ {1} = \mathrm {n} _ {2}$$ $$6 / 2 4 = 0. 2 5 \mathrm {m o l}$$ $$\mathrm {V} = 2 2. 4 * 0. 2 5 = \mathbf {5 . 6 L}$$