# chemistry

posted by .

What weight of na2co3 of 95% purity would be required to neutralize 45.6ml of 0.235n acid?

I want you step by step clearly explain

• chemistry -

mL acid x N acid = milliequivalents acid = 45.6 mL x 0.235 = ? m.e. acid
1 m.e. of anything = 1 m.e. of anything else; therefore, m.e. acid = m.e. Na2CO3.
?m.e. acid = ? m.e. Na2CO3.
m.e. Na2CO3 x milliequivalent weight Na2CO3 = grams Na2CO3
?m.e. Na2CO3 x 0.053 = grams Na2CO3.
Put all of this together to make
mL x N x mew = grams
45.6mL x 0.235N x 0.053g/m.e. =0.56795 g of 100% Na2CO3. Since it is only 95%, then
0.56795/0.95 = 0.5978g which rounds to 0.598g to three significant figures.

## Similar Questions

1. ### chemistry

what weight of na2c03 of 95% purity would be required to neutralize 45.6ml of 0.235N acid?
2. ### math (Damon) clearly step by step for me

3.2g of a mixture of kn03 and nan03 was heated to constant weight which was found to be 2.64g. What is the % kn03 in mixture?
3. ### chemistry Step for step for me thank a lot

A sample of peanut oil weighing 2g is added to 25ml of 0.4M koh. After saponification is complete, 8.5ml of 0.28M hes04 is needed to neutralize excess of koh. The saponification number of peanut oil I do not still understand Pls clearly …
4. ### chemistry step by step for me thank a lot

What is the weight of ca(oh)2 required for 10 litre of water remove temporary hardness of 100 ppm due to ca(H03)2 The answer 0.74 Thank all tutors for your help Pls show step by step for me
5. ### chemistry step by step for me clearly

Al2(s04)3 solution of a molal concentration is present in 1 litre solution of 2.684g/cc. How many moles of bas04 would be precipitated on adding bacl2 in excess. The answer 6M I want some tutors to explain for me step by step clearly. …
6. ### Chemistry step by step thank a lot

10L of hard water required 5.6g of lime for removing hardness. Hence temporary hardness in ppm of cac03 The answer 1000 Who helps me to solve it for me step by step clearly Thank a lot all tutors
7. ### chemistry step by step clearly

10L of hard water required 5.6g of lime for removing hardness. hence temporary hardness in ppm of cac03. The answer 1000ppm I really want all tutors to solve it form clearly step by step. Thank a lot
8. ### chemistry clearly step by step

Al2(s04)3 solution of 1 molal concentration is present in 1 litre solution of 2.684g/cc.How many moles of bas04 would be precipitation on adding bacl2 in excess?
9. ### Chemistry

26.0246 g of sodium carbonate is dissolved in water and the solution volume adjusted to 250.0 ml in a volumetric flask. A 25.00 ml sample of an unknown hydrochloric acid solution required 32.06 ml of the sodium carbonate solution for …
10. ### chemistry

I need help. please show step by step? How many milliliters of 1.0 M NaOH would be required to completely neutralize 40.0 mL of 0.60 M HCl?

More Similar Questions