This is due at 2:00 am today. I only have a few points. No links please I need a real answer. NaOH + AlCl3 --> Al(OH)3 + NaCI.
How many grams of AlCl3 are needed to completely react with 2.25 of NaOH?

7.5 g of AlCl3

Explanation:

The given equation is;

NaOH + AlCl3 --> Al(OH)3 + NaCI.

By inspection, it is not balanced because OH and Clare not equal on both sides of the equation.

Thus, let's make them equal by balancing the equation.

Cl has 3 on the left, so we will make it to have 3 on the right. Same thing with OH on the right and we will make it to have 3 on the left. Thus:

3NaOH + AlCl3 --> Al(OH)3 + 3NaCI

We can see that;

NaOH has 3 moles

While AlCl3 has 1 mole

Thus, to find how many grams of AlCl3 will be required to completely react with 2.25g of NaOH ;

2.25g of NaOH × (3 moles NaOH/39.997 g/mol of NaOH) × (1 mole of AlCl3/3 moles of NaOH) × (133.34 g/mol of AlCl3/1 mol AlCl3) = 7.5 g of AlCl3

jdjxnbfb

explanation:

jjfnc fdnxnxx

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer