Question

question 18 using the equation 2h3po4 + 3mg(oh)2 → mg3(po4)2 + 6h2o, blank 1 grams of water would be produced from 3.57 grams of magnesium hydroxide. blank 1 add your answer 5 points

Explanation:

Step1: Calculate molar mass of Mg(OH)₂

The molar mass of Mg(OH)₂: Mg has an atomic mass of approximately 24.31 g/mol, O has 16 g/mol and H has 1.01 g/mol. So molar mass of Mg(OH)₂ = 24.31+(2×(16 + 1.01))= 58.33 g/mol.

Step2: Calculate moles of Mg(OH)₂

Moles of Mg(OH)₂ = $\frac{mass}{molar\ mass}$. Given mass of Mg(OH)₂ is 3.57 g. So moles of Mg(OH)₂ = $\frac{3.57}{58.33}$ mol.

Step3: Use mole - ratio from balanced equation

From the balanced equation 2H₃PO₄ + 3Mg(OH)₂→Mg₃(PO₄)₂ + 6H₂O, the mole - ratio of Mg(OH)₂ to H₂O is 3:6 or 1:2.

Step4: Calculate moles of H₂O

Moles of H₂O = 2× moles of Mg(OH)₂ = 2×$\frac{3.57}{58.33}$ mol.

Step5: Calculate mass of H₂O

The molar mass of H₂O is 18.02 g/mol. Mass of H₂O = moles of H₂O×molar mass of H₂O = 2×$\frac{3.57}{58.33}$×18.02 g.
\[

$$\begin{align*} \text{Mass of }H_2O&=2\times\frac{3.57}{58.33}\times18.02\\ &=\frac{2\times3.57\times18.02}{58.33}\\ &=\frac{128.6868}{58.33}\\ &\approx2.21\ g \end{align*}$$

\]

Answer:

2.21