b. 2.500a. 3.000j. 47.00s. 262.87aa. 10³ bb. cc. 30.62k. 5.00c. 10.00I. 50.00t. 0.1800d. 1.000e. 2.000m. 25.0 n. 0.02041u. 3.000 x 10³ v. 7.500dd. 1971 ee. 0.01000f. 9.000o. 97.98W. 0.6667ff. 20.41g. 6.000p. 30.62x. 0.200h. 8.000q. 6.572y. 2.67i. 0.02500r. 0.06124z. 2.67 x 10³ How many mL of 3.000 M Mg(OH)2 solution are needed to react with 2.000 L of 3.000 M H3PO4?6 H₂O + Mg3(PO4)22 H3PO4 + 3 Mg(OH)2This is a titration problem, and we are given enough information to calculate the moles of the acid.mol(2.000 L)(-Lmol H3PO4)(-mol Mg(OH)₂)(-L NaOH(-mol Mg(OH)₂mol H₂PO4mol H₂PO4mol Mg(OH)₂mLLumol Mg(OH)2L NaOHmL NaOH

Molarity of MgOH2 solution = 3.00 M Molarity of H3PO4 = 3.00 M Volume of H3PO4 = 2.000 L The given…

Answered: How many mL of 3.000 M Mg(OH)₂ solution…

How many mL of 3.000 M Mg(OH)₂ solution are needed to react with 2.000 L of 3.000 M H3PO4? 2 H3PO4 + 3 Mg(OH)2 6 H₂O + Mg3(PO4)2

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Concept explainers

Solutions

Solutions can be considered as homogeneous mixtures which contain two or more than two chemical substances. The chemical substance which is the greater part is defined as solvent. And the chemical substance that is dissolved in a solvent is referred to as solute. Solution chemistry plays an important role in chemistry since it supports many commercial applications of solutions.

Question

b. 2.500
a. 3.000
j. 47.00
s. 262.87
aa. 10³ bb. cc. 30.62
k. 5.00
c. 10.00
I. 50.00
t. 0.1800
d. 1.000
e. 2.000
m. 25.0 n. 0.02041
u. 3.000 x 10³ v. 7.500
dd. 1971 ee. 0.01000
f. 9.000
o. 97.98
W. 0.6667
ff. 20.41
g. 6.000
p. 30.62
x. 0.200
h. 8.000
q. 6.572
y. 2.67
i. 0.02500
r. 0.06124
z. 2.67 x 10³

How many mL of 3.000 M Mg(OH)2 solution are needed to react with 2.000 L of 3.000 M H3PO4?
6 H₂O + Mg3(PO4)2
2 H3PO4 + 3 Mg(OH)2
This is a titration problem, and we are given enough information to calculate the moles of the acid.
mol
(2.000 L)(-
L
mol H3PO4)(-
mol Mg(OH)₂)(-
L NaOH(-
mol Mg(OH)₂
mol H₂PO4
mol H₂PO4
mol Mg(OH)₂
mL
L
u
mol Mg(OH)2
L NaOH
mL NaOH

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