Question

6,5 гр цинка вступила в реакцию с фосфорной кислотой.Найти массу фосфата цинка и объём водорода (если можно в письменной форме)ответ

Answer 1

3Zn + 2H3PO4=Zn3(PO4)2+3H2↑
n(Zn)=m/M=0.1
n(Zn3(PO4)2)=0,03.
m(Zn3(PO4)2)=n•M=10.52
V(H2)=n•Vm=0,672
Но это не точно, ахпха.

Answer 2

Calculation of the Mass of Zinc Phosphate and Volume of Hydrogen Gas

To find the mass of zinc phosphate and the volume of hydrogen gas produced when 6.5 g of zinc reacts with phosphoric acid, we need to consider the balanced chemical equation for the reaction.

The balanced chemical equation for the reaction between zinc and phosphoric acid is as follows:

Zn + H3PO4 → Zn3(PO4)2 + H2

According to the equation, 1 mole of zinc reacts with 1 mole of phosphoric acid to produce 1 mole of zinc phosphate and 1 mole of hydrogen gas.

To find the mass of zinc phosphate, we need to calculate the number of moles of zinc reacted and then use the stoichiometry of the balanced equation to determine the number of moles of zinc phosphate produced.

The molar mass of zinc (Zn) is approximately 65.38 g/mol, and the molar mass of zinc phosphate (Zn3(PO4)2) is approximately 386.11 g/mol.

Let's calculate the number of moles of zinc reacted:

Number of moles of zinc = mass of zinc / molar mass of zinc

Number of moles of zinc = 6.5 g / 65.38 g/mol = 0.0993 mol

According to the balanced equation, 1 mole of zinc reacts to produce 1 mole of zinc phosphate. Therefore, the number of moles of zinc phosphate produced is also 0.0993 mol.

Now, let's calculate the mass of zinc phosphate:

Mass of zinc phosphate = number of moles of zinc phosphate * molar mass of zinc phosphate

Mass of zinc phosphate = 0.0993 mol * 386.11 g/mol = 38.32 g

Therefore, the mass of zinc phosphate produced is approximately 38.32 g.

To find the volume of hydrogen gas produced, we need to use the ideal gas law, which states that the volume of a gas is directly proportional to the number of moles of the gas.

According to the balanced equation, 1 mole of zinc reacts to produce 1 mole of hydrogen gas. Therefore, the number of moles of hydrogen gas produced is also 0.0993 mol.

Now, let's calculate the volume of hydrogen gas:

Volume of hydrogen gas = number of moles of hydrogen gas * molar volume of gas at STP

The molar volume of gas at STP (standard temperature and pressure) is approximately 22.4 L/mol.

Volume of hydrogen gas = 0.0993 mol * 22.4 L/mol = 2.22 L

Therefore, the volume of hydrogen gas produced is approximately 2.22 L.

In summary, when 6.5 g of zinc reacts with phosphoric acid, the mass of zinc phosphate produced is approximately 38.32 g, and the volume of hydrogen gas produced is approximately 2.22 L.

Please note that the values calculated here are approximate and may vary slightly depending on the actual molar masses and experimental conditions.