8 Answered Questions for the topic Stoichiometery

Stoichiometery

09/18/18

#### 2 V2O5(s) + 5 Si(l) → 4 V(l) + 5 SiO2(s) Given the above balanced equation, calculate the number of grams of silicon required to prepare 359 grams of van

2 V2O5(s) + 5 Si(l) → 4 V(l) + 5 SiO2(s)Given the above balanced equation, calculate the number of grams of silicon required to prepare 359 grams of vanadium metal.Your answer should have 3 sig figs.

Stoichiometery

01/22/18

#### How many hydrogen atoms would be in 13.7 of CuSO4*5H2O

show the formula’s used

Stoichiometery

12/18/17

#### If 38.0 ml of 1.25 M sulphuric acid solution reacts with a 27.5 ml of 0.5 M solution containing BaCl2

a write a balanced chemical equation
b determine the theoretical yield of solid in reaction
c if 2.5 g of was actually produce, determine percent yield of reaction

Stoichiometery Concentration Of Solutions

11/27/17

#### chemistry from stoichiometry

1kg of 1m glucose solution is diluted to 5kg the molality of the diluted solution should be

Stoichiometery

04/04/17

#### The decomposition of potassium chlorate yields oxygen gas. If the yield is 65%, how many grams of KClO3 are needed to produce 47.5L of O2?

chemistry problem

03/27/17

#### Got a Chemistry Stoichiometry Problem.. Please Help!

Question: The final step in manufacture of pure platinum (for use in automobile catalytic converters and other purposes) is the decomposition reaction of (NH4)2PtCl6. In addition to the solid...
more

09/21/15

#### Chemistry with stoichiometry and empirical formula

A compound that does not contain oxygen is combusted to give: a) 33.75 g of water b) Carbon dioxide collected under water in a 37.5 L vessel at 20 degrees Celsius and an atmospheric pressure of...
more

Stoichiometery

01/08/14

#### given 97.8 grams of s'more , how many graham crackers will you need

how should i solve this problem. should i got to use the formula mass or should i used molecular formula

## Still looking for help? Get the right answer, fast.

Ask a question for free

Get a free answer to a quick problem.

Most questions answered within 4 hours.

#### OR

Find an Online Tutor Now

Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.